Trending ▼   ResFinder  

CBSE 10th Board Class 10 2020 : Mathematics

28 pages, 29 questions, 0 questions with responses, 0 total responses,    0    0
Sumayya Sumi
   		+Fave Message
ProfileTimelineUploads
 Home > sumayya35 >

Formatting page ...

Chap 9 : Some Applications of Trigonometry www.cbse.online CHAPTER 9 Some Applications of Trigonometry twice the distance between the foot of the ladder and the wall. Find the angle made by the ladder with the horizontal. Ans : [CBSE 2015, Set-HODM40L] VERY SHORT ANSWER TYPE QUESTIONS 1. A ladder 15 m long leans against a wall making an angle of 60 with the wall. Find the height of the point where the ladder touches the wall. Ans : [KVS 2014] Let the distance between the foot of the ladder and the wall is x , then length of the ladder will be 2x . As per given in question we have drawn figure below. Let the height of wall be h . As per given in question we have drawn figure below. h = cos 60 15 In TABC, h = 15 # cos 60 = 15 # 1 = 7.5 m 2 2. A pole casts a shadow of length 2 3 m on the ground, when the Sun s elevation is 60 . Find the height of the pole. Ans : [CBSE Foreign 2015] Let the height of pole be h. As per given in question we have drawn figure below. h = tan 60 2 3 h = 2 3 tan 60 =2 3# 3. 3 =6 m If the length of the ladder placed against a wall is +B = 90 cos A = x = 1 = cos 60 2x 2 A = 60 Add 8905629969 in Your Class Whatsapp Group to Get All PDFs 4. An observer, 1.7 m tall, is 20 3 m away from a tower. The angle of elevation from the eye of observer to the top of tower is 30 . Find the height of tower. Ans : [CBSE Foreign 2016] Let height of the tower AB be h . As per given in question we have drawn figure below. Here AE = h - 1.7 and BC = DE = 20 3 In TADE, +E = 90 tan 30 = h - 1.7 20 3 Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 172 Chap 9 : Some Applications of Trigonometry www.rava.org.in 1 = h - 1.7 3 20 3 m away from the wall, find the length of the ladder. Ans : [CBSE Board Term-2, 2011] h - 1.7 = 20 or 5. Let the length of ladder be x . As per given in question we have drawn figure below. h = 20 + 1.7 = 21.7 m In figure, a tower AB is 20 m high and BC, its shadow on the ground, is 20 3 m long. find the Sun s altitude. In TACB with +C = 60c cos 60c = 2.5 AC [CBSE Outside Delhi 2015] Ans : 1 = 2.5 2 AC Let the +ACB be q . tan q = AB = 20 = 1 = tan 30c BC 20 3 3 Thus 6. For more files visit www.cbse.online q = 30c In the given figure, AB is a 6 m high pole and DC is a ladder inclined at an angle of 60c to the horizontal and reaches up to point D of pole. If AD = 2.54 m, find the length of ladder. ( use 3 = 1.73 ) 8. As per given in question we have drawn figure below. AD = 2.54 m DB = 6 - 2.54 = 3.46 m In TBCD , Here AE = 1.5 m is height of observer and BD = 30 m is tower. +B = 90c sin 60c = BD DC Now In TBAC, 3 = 3.46 2 DC q = 45c Thus length of ladder is 4 m. A ladder, leaning against a wall, makes an angle of 60c with the horizontal. If the foot of the ladder is 2.5 BC = 30 - 1.5 = 28.5 m tan q = BC AC tan q = 28.5 = 1 = tan 45c 28.5 DC = 3.46 # 2 = 3.46 = 4 1.73 3 7. An observer 1.5 m tall is 28.5 m away from a tower 30 m high. Find the angle of elevation of the top of the tower from his eye. Ans : [CBSE Board Term-2, 2012] [CBSE Delhi 2016] Ans : We have AC = 2 # 2.5 = 5 m Hence angle of elevation is 45c 9. If the angles of elevation of the top of a tower from two points distant a and b ^a 2 b h from its foot and in Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 173 Chap 9 : Some Applications of Trigonometry www.cbse.online 1 = 150 d 3 the same straight line from it are respectively 30c and 60c, then find the height of the tower. Ans : [CBSE 2014] Let the height of tower be h . As per given in question we have drawn figure below. From TABD, From TACD, From (1) From (2) Thus d = 150 3 m. Thus 11. In the given figure, if AD = 7 3 m, then find the value of BC . [CBSE 2012] Ans : h = tan 30c a Let BD = x and DC = y h =a# 1 = a 3 3 h = tan 60c b ...(1) h = b# ...(2) 3 =b 3 From TABD we get tan 30c = 7 3 x 1 =7 3 x 3 a = 3h b = h 3 a#b = x =7 3# From TADC , tan 60c = 7 3 y 3h# h 3 3 =7 3 y ab = h2 h = ab Hence, the height of the tower is 3 = 21 m ab . 10. The angle of depression of a car parked on the road from the top of a 150 m high tower is 30c. Find the distance of the car from the tower (in m). Ans : [CBSE Outside Delhi, 2014] Let the distance of the car from the tower be d . As per given in question we have drawn figure below. y = 7 m. Now BC = BD + DC = 21 + 7 = 28 m. Hence, the value of BC is 28 m. 12. The top of two poles of height 16 m and 10 m are connected by a length l meter. If wire makes an angle of 30c with the horizontal, then find l . Ans : [CBSE Board Term-2, 2012] Let BD and AE be two poles, where BD = 16 m, AE = 10 m. As per given in question we have drawn figure below. Due to alternate angles we have +BAX = +ABC = 30c In TACB , +C = 90c tan 30c = 150 d Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 174 Chap 9 : Some Applications of Trigonometry Length www.rava.org.in BC = BD - CD = BD - AE sin 60c = AB BC = 16 - 10 = 6 m. From TABC , sin 30c = BC l 3 = 100 2 l l = 2 # 100 = 200 m 3 3 1 = BC 2 l = 200 # 3 l = 2BC = 6 # 2 = 12 m. 3 = 200 3 m 3 3 Hence length the kids string is 200 3 3 Hence, the value of l is 12 m. 13. A pole 6 m high casts a shadow 2 3 m long on the ground, then find the Sun s elevation. Ans : [CBSE Board Term-2, 2012 ] Let the Sun s elevation be q . As per given in question we have drawn figure below. 15. Find the angle of elevation of the top of the tower from the point on the ground which is 30 m away from the foot of the tower of height 10 3 m. Ans : [CBSE Board Term-2, 2012] Let the angle of elevation of top of the tower be q . As per given in question we have drawn figure below. From TABC , Length of pole is 6 m and length of shadow is 2 3 m. tan q = AB = 10 3 = 1 = tan 30c 30 BC 3 From TABC, we have Thus tan q = AB BC 6 tan q = = 3 = 2 3 3 q = 30c Hence angle of elevation is 30c. 3 = tan 60c q = 60c Hence sun s elevation is 60c. 14. Find the length of kite string flying at 100 m above 16. If the altitude of the sun is 60c, what is the height of a tower which casts a shadow of length 30 m ? Ans : [CBSE Board Term-2, 2011] Let AB be the tower whose height be h m. As per given in question we have drawn figure below. the ground with the elevation of 60c. Ans : [CBSE Board Term-2, 2012] Let the length of kite string AC = l m. As per given in question we have drawn figure below. Here shadow is BC = 30 m. From TABC , we get AB = tan 60c BC Here +ACB = 60c, height of kite AB = 100 m. From TABC , we have h = 30 3 h = 30 3 m Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 175 Chap 9 : Some Applications of Trigonometry www.cbse.online far the boat from the base of the light house. Ans : [CBSE Board Term-2, 2015] Hence, height of tower is 30 3 m. 17. If cos A = 2 , find the value of 4 + 4 tan2 A . 5 Ans : [CBSE Sample Question Paper 2017-18] Let AB be the light house and C be the position of the boat. As per given in question we have drawn figure below. 4 +| 4 tan2 A = 4 ^1 + tan2 Ah 4 sec2 A = 42 = 24 2 = 4 # 25 = 25 4 cos A ^5h 18. The ratio of the height of a tower and the length of its shadow on the grow is elevation of the sun ? Ans : 3 | 1. What is the angle of [CBSE Board Term-2, 2016] Let height of tower be AB and its shadow be BC . As per given in question we have drawn figure below. Since +PAC = 60c & +ACB = 60c Let In TABC , CB = x tan 60c = AB BC 3 = 40 x x = 40 = 40 # 3 = 40 3 m 3 3 3# 3 Hence, The boat is 40 3 m away from the foot of 3 light house. AB = tan q = 3 = tan 60c 1 BC Hence, angle of elevation of sun is 60c. For more files visit www.cbse.online 19. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of the sun ? [CBSE Outside Delhi 2017] Ans : Tower AB is 30 m and shadow BC is 10 3 . As per given in question we have drawn figure below. 2. A kite is flying at a height of 90 m above the ground. The string attached to the kite is temporarily tried to a point on the ground. The inclination of the string with the ground is 60c. Find the length of the string assuming that there is no slack in the string. Ans : [CBSE Delhi Term-2, 2014, 2011] As per given in question we have drawn figure below. In TABC which is right triangle, tan q = AB = 30 = 3 = tan 60c BC 10 3 Thus q = 60c so, angle of elevation of sun is 60c. SHORT ANSWER TYPE QUESTIONS - I 1. From the top of light house, 40 m above the water, the angle of depression of a small boat is 60c. Find how In right TABC , we have sin 60c = AB AC 3 = 90 x 2 x = 90 # 2 = 180 = 3 # 60 3 3 3 = 60 3 = 60 # 1.732 Hence length of string is 103.92 m. Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 176 Chap 9 : Some Applications of Trigonometry 3. www.rava.org.in A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30c with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. Ans : [CBSE Board Term-2, 2011, Set A1] In TABC , Let the tree be AC and is broken at B . The broken part touches at the point D on the ground. As per given in question we have drawn figure below. In TABD , AB = tan 30c BC AB = tan 30c = 1 30 3 AB = 30 = 10 3 3 AB = tan 60c BD 10 3 = tan 60c = BD 3 BD = 10 m Hence the length of shadow is 10 m. 5. From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30c. A flag is hoisted at the top the of the building and the angle of elevation of the length of the flagstaff from P is 45c. Find the length of the flagstaff and distance of building from point P . [Take 3 = 1.732 ] Ans : [ CBSE Board Term-2, 2012] [Delhi 2013] [ Term-2, 2011] Let height of flagstaff be BD = x m. As per given in question we have drawn figure below. In right TBCD , cos 30c = CD BD 3 = 8 2 BD BD = 16 3 BC tan 30c = CD and 1 = BC 8 3 BC = 8 3 tan 30c = AB AP Height of tree, 1 = 10 AP 3 BC + BD = 8 + 16 = 24 = 8 3 3 3 3 AP = 10 3 Hence the height of the tree is 8 3 m. 4. Distance of the building from P , If the shadow of a tower is 30 m long, when the Sun s elevation is 30c. What is the length of the shadow, when Sun s elevation is 60c ? Ans : [CBSE Board Term-2, 2011, Set C1] Now = 10 # 1.732 = 17.32 m tan 45c = AD AP 1 = 10 + x 17.32 As per given in question we have drawn figure below. x = 17.32 - 10.00 = 7.32 m Hence, length of flagstaff is 7.32 m. 6. The angle of elevation of the top of a building from the foot of the tower is 30c and the angle of elevation of the top of the tower from the foot of the building is 45c. If the tower is 30 m high, find the height of the building. Ans : [Delhi CBSE 2015 Set I, II, III] Let the height of the building be AB = h m. and distant between tower and building be, BD = x m. As per given in question we have drawn figure below. Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 177 Chap 9 : Some Applications of Trigonometry www.cbse.online SHORT ANSWER TYPE QUESTIONS - II 1. An electric pole is 10 m high. A steel wire tied to top of the pole is affixed at a point on the ground to keep the pole up right. If the wire makes an angle of 45 with the horizontal through the foot of the pole, find the length of the wire. [Use 2 = 1.414 ] Ans : [CBSE Term 2, 2016] Let OA be the electric pole and B be the point on the ground to fix the pole. Let BA be x . As per given in question we have drawn figure below. In TABD tan 45c = AB BD 1 = 30 x x = 30 ...(1) Now in TBDC , tan 30c = CD BD 1 =h x 3 3h = x & h = x 3 In TABO, we have ...(2) sin 45 = AO AB From (1) and (2), we get h = 30 = 10 3 m. 3 1 = 10 x 2 Therefore height of the building is 10 3 m 7. A player sitting the top of a tower of height 20 m observes the angle of depression of a ball lying on the ground as 60c. Find the distance between the foot of the tower and the ball. Take 3 = 1.732 Ans : [CBSE Board Term-2, 2011, B1] Let C be the point where the ball is lying. As per given in question we have drawn figure below. x = 10 2 = 10 # 1.414 = 14.14 m Hence, the length of wire is 14.14 m 2. The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45 and 60 respectively. Find the height of the tower and the horizontal distance between the tower and the building. (Use 3 = 1.73) Ans : [Delhi Set I, II, III, 2016] As per given in question we have drawn figure below. Due to alternate angles we obtain +XAC = +ACB = 60c In TABC , tan 60c = AB BC 3 = 20 x x = 20 = 20 c 3 m 3 3 Hence, distance between ball and foot of tower is 11.53 m. We have and tan 45 = h - 50 x x = h - 50 tan 60 = h x Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in ...(1) Page 178 Chap 9 : Some Applications of Trigonometry www.rava.org.in 3 =h x x = h 3 of the flagstaff are 60 and 45 respectively. Find the height of the tower correct to one place of decimal. (Use 3 = 1.73) Ans : [CBSE Foreign Set II, 2016] ...(2) As per given in question we have drawn figure below. From (1) and (2) we hav h - 50 = h 3 3 h - 50 3 = h 3 h - h = 50 3 h ^ 3 - 1h = 50 3 50 ^3 + h = 50 3 = 2 3 -1 h = 25 (3 + 3h 3) = 75 + 25 3 = 118.25 m 3. An aeroplane, when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are 60 and 45 respectively. Find the vertical distance between the aeroplanes at that instant. (Use 3 = 1.73) Ans : [Foreign Set III, 2016] Let the height first plane be AB = 4000 m and the height of second plane be BC = x m. As per given in question we have drawn figure below. x = tan 45 = 1 & x = y y x + 7 = tan 60 = x 3 7 = ^ 3 - 1h x x = 5. 7 ^ 3 + 1h 7 (2.73) = 9.6 m = 2 2 Two men on either side of a 75 m high building and in line with base of building observe the angles of elevation of the top of the building as 30 and 60 . find the distance between the two men. (Use 3 = 1.73) Ans : [CBSE Foreign Set I, 2016] Let AB be the building and the two men are at P and Q. As per given in question we have drawn figure below. Here +BDC = +45 and +ADB = 60 x = tan 45 = 1 & x = y In TCBD , y and in TABD , 4000 = tan 60 = y 3 y = 4000 3 3 In TABP , 1 = 75 BP 3 = 2306.67 m Thus vertical distance between two, 4000 - y = 4000 - 2306.67 = 1693.33 m 4. A 7 m long flagstaff is fixed on the top of a tower standing on the horizontal plane. From point on the ground, the angles of elevation of the top and bottom tan 30 = AB BP In TABQ, BP = 75 3 m tan 60 = AB BQ 3 = 75 BQ BQ = 75 = 25 3 3 Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 179 Chap 9 : Some Applications of Trigonometry www.cbse.online Distance between the two men, AD = BC = 3000 3 m PQ = BP + BQ = 75 3 + 25 3 Let the speed of the aeroplane = x m/s = 100 3 = 100 # 1.73 = 173 6. AB = DC # 30 # x = 30x m ...(1) The horizontal distance between two towers is 60 m. The angle of elevation of the top of the taller tower as seen from the top of the shorter one is 30 . If the height of the taller tower is 150 m, then find the height of the shorter tower. Ans : [CBSE Board Term-2, 2015] In right TAED, we have tan 60 = AD DE Let AB and CD be two towers. Let the height of the shorter tower AB = h m. As per given in question we have drawn figure below. In right TBEC, 3 = 3000 3 DE DE = 3000 m ...(2) tan 30 = BC EC 1 = 3000 3 DE + CD 3 DE + CD = 3000 # 3 3000 + 30x = 9000 30x = 6000 x = 200 m/s Hence, Speed of plane is 200 m/s = 200 # 18 = 720 km/hr 5 8. Here BC = AE = 60 m, DE = DC - EC = (150 - h) DE = tan 30 In TAED , AE 150 - h = tan 30 = 1 60 3 A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60 and the angle of depression of the base of hill as 30 . Find the distance of the hill from the ship and the height of the hill. Ans : [Outside Delhi, Set-II, 2016] As per given in question we have drawn figure below. Here AC is height of hill and man is at E . ED = 10 is height of ship from water level. As per given in question we have drawn figure below. 150 3 - h 3 = 60 3 h = 150 3 - 60 3 h = 150 3 - 20 3 # or 7. 3 h = ^150 - 20 3 h m The angle of elevation of an aeroplane from a point on the ground is 60 . After a flight of 30 seconds the angle of elevation becomes 30 . If the aeroplane is flying at a constant height of 3000 3 m, find the speed of the aeroplane. Ans : [CBSE O.D. 2014] As per given in question we have drawn figure below. In TBCE, BC = 10 m and Now +BEC = 30 tan 30 = BC BE 1 = 10 BE 3 +AED = 60 , +BED = 30 BE = 10 3 Since BE = CD , distance of hill from ship CD = 10 3 m = 10 # 1.732 m Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 180 Chap 9 : Some Applications of Trigonometry www.rava.org.in = 17.32 m building are 30 and 45 , respectively. Find the height of multi-storeyed building and distance between two buildings. [KVS 2014] Ans : Now in TABE , +AEB = 60 where AB = hm, BE = 10 3 m and Thus +AEB = 60 tan 60 = AB BE As per given in question we have drawn figure below. 3 = AB 10 3 AB = 10 3 # 3 = 30 m Thus height of hill AB + 10 = 40 m 9. Two ships are approaching a light house from opposite directions. The angle of depression of two ships from top of the light house are 30 and 45 . If the distance between two ships is 100 m, Find the height of lighthouse. Ans : [CBSE Foreign 2014] As per given in question we have drawn figure below. Here AD is light house of height h and BC is the distance between two ships. Here AE = CD = 8 m BE = AB - AE = ^h - 8h m and AC = DE = x m Also, +FBD = +BDE = 30c +FBC = +BCA = 45c In right angled TCAB we have tan 45c = AB AC 1 =h & x=h x ...(1) In right angled TEDB We have In TADC In TABD, tan 30c = BE ED BC = 100 m tan 45 = h & h = x x tan 30 = 1 = h-8 x 3 h 100 - x x = h 1 = 100 - x 3 ...(2) From (1) and (2), we get h = 100 - x = h 3 100 - h = h 3 3 ^h - 8h 8 3 = h=x = h ^1 + 3 h h = 100 1+ 3 = 50 # 0.732 Thus height of light house is 36.60 m. 10. The angles of depression of the top and bottom of an 8 m tall building from top of a multi-storeyed 3 +1 3 +1 = 4 3 ^ 3 + 1h = ^12 + 4 3 h m Since, x = h , ^ 3 - 1h = 100 # ^ 3 + 1h ^ 3 - 1h 100 ^ 3 - 1h = 3-1 = 50 ^1.732 - 1h 3h-h h = 8 3 # 3 -1 100 = h + h 3 = 50 ^ 3 - 1h 3h-8 3 x = ^12 + 4 3 h Distance = ^12 + 4 3 h m Hence the height of multi storey building = distance = ^4 3 + 12h m. 11. From a top of a building 100 m high the angle of depression of two objects are on the same side observed to be 45c and 60c. Find the distance between the objects. Ans : [CBSE Board Term-2, 2014] Let A be a point on top of building and B , C be two objects. As per given in question we have drawn figure Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 181 Chap 9 : Some Applications of Trigonometry www.cbse.online below. AB = = In TABO , 2 BO + AO 2 ^15k h2 + ^8k h2 = 17k AO = sin q AB h = 15k = 15 90 17k 17 h = 15 # 90 = 79.41 m 17 Hence, height of kite is 79.41 m. 13. Two men standing on opposite sides of a tower Here +ACO = +CAX = 45c +ABO = +XAB = 60c In right TAOC , AO = tan 45c CO and measure the angles of elevation of he top of the tower as 30c and 60c respectively. If the height of the tower in 20 m, then find the distance between the two men. Ans : [CBSE Board Term-2, 2013] Let two men are standing at A and C and BT is the tower. As per given in question we have drawn figure below. 100 = 1 CO CO = 100 m Also in right TAOB , AO = tan 60c OB 100 = 3 OB OB = 100 3 Thus BC = CO - OB = 100 - 100 3 ^ 3 - 1h = 100 c1 - 1 m = 100 3 3 ^ 3 - 1h 3 # 3 3 100 ^3 - 3 h m = 3 12. A Boy, flying a kite with a string of 90 m long, which is making an angle q with the ground. Find the height of the kite. (Given tan q = 45 ) 8 Ans : [CBSE Board Term-2, 2014] = 100 Let A be the position of kite and AB be the string. As per given in question we have drawn figure below. In right angle triangle TBTC , tan 60c = BT BC 1 = 20 BC 3 BC = 20 3 In right angle triangle TBTC, tan 60 = BT BC 3 = 20 BC BC = 20 3 Thus distance between two men AB + BC = 20 3 + 20 = 60 + 20 = 80 3 m. 3 3 3 Hence, distance between the men is 80 3 m. 3 14. Two poles of equal heights are standing opposite to Since tan q = 15 = AO 8 BO Let AO be 15k and BO be 8k Now using Pythagoras Theorem each other on either side of a road, which is 80 m wide. From a point between them on the road, angles of elevation of their top are 30c and 60c. Find the height of the poles and distance of point from poles. Ans : [CBSE Board Term-2, 2011 Set (B1), Delhi 2013] Let the distance between pole AB and man E be x . As per given in question we have drawn figure below. Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 182 Chap 9 : Some Applications of Trigonometry www.rava.org.in In right TPTR , tan 30c = PT TR 1 = h 15 3 h = 15 = 5 3 3 = 5 # 1.732 = 8.66 PQ = PT + TQ = 8.66 + 24 Here distance between pole CD and man is 80 - x In right angle triangle TABE , tan 30c = h x h = x 3 = 32.66 m Thus height of the second pole is 32.66 m. 16. The angle of elevation of the top of a tower from a point A on the ground is 30c. On moving a distance of 20 metre towards the foot of the tower to a point B the angle of elevation increase to 60c. Find the height of the tower and the distance of the tower from the point A. Ans : [CBSE Board Term-2, 2012] ...(1) In angle triangle TCDE , tan 60c = h 80 - x 3 = h 80 - x h = 80 3 - x 3 Let height of tower be h and distance BC be x . As per given in question we have drawn figure below. ...(2) Comparing (1) and (2) we have x = 80 3 - x 3 3 x = 80 # 3 - x # 3 4x = 240 x = 240 = 60 m 4 Substituting this value of x in (1) we have h = 60 = 20 3 3 Hence, height of the pole is 34.64 m In right TDBC , h = tan 60c x 15. The horizontal distance between two poles is 15 m. The angle of depression of the top of first pole as seen from the top of second pole is 30c. If the height of the first of the pole is 24 m, find the height of the second pole. [ Use 3 = 1.732 ] Ans : [CBSE Board Term-2, 2013] h = 3x ...(1) In right TADC , h = tan 30c = 1 x + 20 3 ...(2) 3 h = x + 20 Substituting the value of h from eq. (1) in eq. (2), we get Let RS be first pole and PQ be second pole. As per given in question we have drawn figure below. 3x = x + 20 x = 10 m Thus and ...(3) AC = 20 + x = = 30 m. h = 3 # 10 = 10 3 = 10 # 1.732 = 17.32 m Hence, height of tower is 17.32 m and distance of tower from point A is 30 m. 17. The angle of elevation of the top of a hill at the foot of a tower is 60c and the angle of elevation of the top of the tower from the foot of the holl is 30c. If the tower is 50 m high, find the height of the hill. Ans : [CBSE Board Term-2, 2012] Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 183 Chap 9 : Some Applications of Trigonometry www.cbse.online Let AB be tower of height 50 m and DC be hill of height h . As per given in question we have drawn figure below. In right TDCA , h = tan 30c = 1 x + 50 3 3 h = x + 50 (1) Substituting the value of h from (1) in (2), we have 3x = x + 50 2x = 50 & x = 25 m h = 25 3 = 25 # 1.732 = 43.3 m Hence height of tower is 43.3 m. 19. A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground the angle of elevation of the top of the statue is 60c and from the same point the angle of elevation of the top of the pedestal is 45c . Find the height of the pedestal. Ans : [CBSE Board Term-2, 2012 (50)] In right TBAC cos 30c = AC 50 Let CD be statue of 1.6 m and pedestal BC of height h . Let A be point on ground. As per given in question we have drawn figure below. 3 = AC 50 AC = 50 3 In right TACD , tan 60c = CD 50 3 3 = CD 50 3 CD = 50 3 # 3 = 150 m Thus height of the hill CD = 150 m 18. A person observed the angle of elevation of the top of a tower as 30c. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60c. Find the height of the tower. Ans : [CBSE Board Term-2, 2012 Set(31)] Let DC be tower of height h . As per given in question we have drawn figure below. In right TABD , cos 60c = AB BD 1 = AB h + 1.6 3 AB = h + 1.6 3 ...(1) In right TABC , AB = cot 45c BC 1 = AB h AB = h Here A is the point at elevation 30c and B is the point of elevation at 60c Let BC be x . AC = ^50 + x h m In right TDCB , h = tan 60c = 3 x Now h = 3x ...(2) From (1) and (2), we get h = h + 1. 6 3 h 3 = h + 1.6 h 3 - h = 1.6 h ^ 3 - 1h = 1.6 ...(1) h = 1.6 1.6 = 1.732 - 1 3 -1 Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 184 Chap 9 : Some Applications of Trigonometry www.rava.org.in = 1.6 = 2.185 m 0.732 Height of pedestal h is 2.2 m. 20. From a point on a ground, the angle of elevation of bottom and top a transmission tower fixed on the top of a 20 m high building are 45c and 60c respectively. Find the height of the tower. Ans : [Outside Delhi Compt. 2017] Let P be the point on ground, AB be the building of height 20 m and BC be the tower of height x . As per given in question we have drawn figure below. In right TBAC , AB = tan 60c AC h = x 3 h =x 3 In right TBAD , AB = tan q AD h = tan q 3x x 3 = 1 = tan 30c 3x 3 Thus q = 30c. In right TBAP we have BA = tan 45c PA 20 = 1 y y = 20 In right TCAP CA = tan 60c PA 20 + x = y 22. On a straight line passing through the foot of a tower, two C and D are at distance of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower. Ans : [CBSE Outside Delhi 2017] Let AB be tower of height h , C and D be the two point. As per given in question we have drawn figure below. 3 20 + x = y 3 20 + x = 20 3 x = 20 3 - 20 = 20 ^ 3 - 1h = 20 # ^1.732 - 1h = 20 # 0.73 = 14.64 Hence, height of the tower is 14.64 m. 21. The shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the sun is 60c. Find the angle of elevation of the sun at the of the longer shadow. Ans : [CBSE Foreign 2017] Let AB be tower of height h , AC be the shadow at elevation of sun of 60c. As per given in question we have drawn figure below. Since +ACB and +ADB are complementary, +ACB = q and +ADB = 90c - q Now, in right TABC , ...(1) tan q = AB = h 4 BC In right TABD , tan ^90 - qh = AB = h BD 16 cot q = h 16 Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 tan _90 - q i = cot q Page 185 Chap 9 : Some Applications of Trigonometry tan q = 16 h www.cbse.online distance XB . Ans : ...(2) [CBSE SA-2 2016] As per given in question we have drawn figure below. From (1) and (2) we have h = 16 4 h h2 = 4 # 16 = 64 = 82 h =8 m Thus height of tower is 18.8 m. LONG ANSWER TYPE QUESTIONS 1. From the top of tower, 100 m high, a man observes two cars on the opposite sides of the tower with the angles of depression 30c & 45c respectively. Find the distance between the cars. (Use 3 = 1.73 ) Ans : [CBSE Board Sample Paper, 2016] In right TYCB , we have tan 45c = BC YC Let DC be tower of height 100 m. A and B be two car on the opposite side of tower. As per given in question we have drawn figure below. 1 = x YC YC = x XA = x In right TXAB we have tan 60c = AB XA 3 = x + 40 x 3 x = x + 40 In right TADC , x 3 - x = 40 tan 30c = CD AD x = 1 = 100 x 3 x = 100 3 = ^20 3 + 1h ...(1) Thus height of the tower, AB = x + 40 tan 45c = CD DB = 20 3 + 20 + 40 1 = 100 y = 20 3 + 60 = 20 ^ 3 + 3h y = 100 m In right TXAB we have, sin 60c = AB BX Distance between two cars AB = AD + DB = ^100 3 + 100h 3 = AB 2 BX = ^100 # 1.73 + 100h m 20 # 2 ^ 3 + 3h BX = 2AB = 3 3 = ^173 + 100h m = 273 m = 40 ^1 + Hence, distance between two cars is 273 m. 2. 3 +1 3 +1 = 20 ^ 3 + 1h In right TBDC , & 40 # 3 -1 The angle of elevation of the top B of a tower AB from a point X on the ground is 60c. At point Y , 40 m vertically above X , the angle of elevation of the top is 45c. Find the height of the tower AB and the 3h = 40 # 2.73 = 109.20 3. A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 5 m. From a point on the ground the angles of elevation of top and bottom of the flagstaff are 60c and 30c respectively. Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 186 Chap 9 : Some Applications of Trigonometry www.rava.org.in Find the height of the tower and the distance of the point from the tower. (take 3 = 1.732 ) Ans : [CBSE Foreign Set I, 2016] Now QT = ^h - 40h m In right TPQX we have, tan 60c = h x Let AB be tower of height x and AC be flag staff of height 5 m. As per given in question we have drawn figure below. 3 =h x h = ...(1) 3x In right TQTY we have tan 45c = h - 40 x 1 = h - 40 x x = h - 40 ...(2) Solving (1) and (2), we get x = 3 x - 40 3 x - x = 40 ^ 3 - 1h x = 40 In right TABP , AB = tan 30c BP Thus x = 40 = 20 3 + 1 m ^ h 3 -1 x = 3 # 20 ^ 3 + 1h = 20 ^3 + x = 1 y 3 y = = 20 ^3 + 1.73h Hence, height of tower is 94.6 m. 3 ...(2) Substituting the value of y from (1) we have x+5 = 3 3x x + 5 = 3x & x = 2.5 m Height of tower is = 2.5 m Distance of P from tower = ^2.5 # 1.732h or 4.33 m. 4. = 20 # 4.73 ...(1) 3x In right TCBP x + 5 = tan 60c = y 3h m 5. Two post are k metre apart and the height of one is double that of the other. If from the mid-point of the line segment joining their feet, an observer finds the angles of elevation of their tops to be complementary, then find the height of the shorted post. Ans : [CBSE Foreign 2015] Let AB and CD be the two posts such that AB = 2 CD . Let M be the mid-point of CA. As per given in question we have drawn figure below. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60c. From a point Y 40 m vertically above X , the angle of elevation of the top Q of tower is 45c. Find the height of the PQ and the distance PX . (Use 3 = 1.73 ) Ans : [CBSE Outside Delhi 2016] Let PX be x and PQ be h . As per given in question we have drawn figure below. Here +CMD = q and +AMB = 90c - q Clearly, CM = MA = 1 k 2 Let CD = h . then AB = 2h AB = tan 90c - q Now, ^ h AM 2h = cot q k 2 4h = cot q k Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 ...(1) Page 187 Chap 9 : Some Applications of Trigonometry www.cbse.online Also in right TCMD , CD = tan q CM 45c, how soon after this, the car will reach the tower ? Ans : [KVS 2014] Let AB be the tower of height h . As per given in question we have drawn figure below. h = tan q k 2 2h = tan q k ...(2) Multiplying (1) and (2), we have 4h 2h = tan q cot q = 1 # k # k 2 h2 = k 8 k =k 2 4 2 2 The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground flagstaff fixed at the top of the tower, at A is 60c, then find the height of the flagstaff. [Use 3 = 1.73 ] Ans : [CBSE OD 2014] h = 6. Car is at P at 30c and is at Q at 45c elevation. Here +AQB = 45c Now, in right TABQ we have, tan 45c = AB BQ Let BD be the tower of height x and CD be flagstaff of height h . As per given in question we have drawn figure below. 1 = h BQ BQ = h In right TAPB we have, tan 30c = AB PB 1 = h x+h 3 x+h = h 3 x = h ^ 3 - 1h h ^ 3 - 1h m/min 12 Time for remaining distance, h h ^ 3 - 1h 12 t = = 12 ^ 3 - 1h Thus, Here +DAB = 45c, +CAB = 60c and AB = 120 m In right angled TABD we have x = tan 45c = 1 AB 12 ^ 3 + 1h 12 ^ 3 + 1h = 3-1 ^ 3 - 1h^ 3 + 1h = 12 ^ 3 + 1h 2 = x = AB = 120 m In right angled TACB we have h + x = tan 60c = 120 = 6 ^ 3 + 1h t = 6 # 2.73 = 16.38 3 h + 120 = 120 3 h = 120 3 - 120 = 120 ^ 3 - 1h = 120 ^1.73 - 1h = 120 # 0.73 h = 87.6 m Hence, height of the flagstaff is 87.6 m. 7. Speed = A man on the top of a vertical tower observes a car moving at a uniform speed towards him. If it takes 12 min. for the angle of depression to change from 30c to Hence, time taken by car is 16.38 minutes. 8. From the top of a building 60 m high the angles of depression of the top and the bottom of a tower are observed to be 30c and 60c. Find the height of the tower. Ans : [Delhi, Term-2 2014], [CBSE Board Term-2 2012 Set 3, 2011 Set B1] Let AB be the building of height 60 m and CD be the tower of height h . Angle of depressions of top and bottom are given 30c and 60c respectively. As per given in question we have drawn figure below. Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 188 Chap 9 : Some Applications of Trigonometry www.rava.org.in In right TABD we have tan 60c = AB BD 3 = 60 x x = 60 = 20 3 3 Now, in right TBCD we have tan 30c = CD = h x BD 1 = h 20 3 3 Here h = 20 3 = 20 3 DC = EB = h m and let BC = x AE = ^60 - h h m In right angled TAED we have 60 - h = tan 30c ED Hence height of the building is 20 m. 10. The angle of elevation of a cloud from a point 120 m above a lake is 30c and the angle of depression of its reflection in the lake is 60c. Find the height of the cloud. Ans : [CBSE Board Term-2, 2012] 60 - h = 1 x 3 3 ^60 - h h = x As per given in question we have drawn figure below. ...(1) In right TABC we have 60 = tan 60c x ...(2) 60 = 3 x Substituting the value of x from equation (1) in equation (2), we have 60 = 3# 3 ^60 - h h 60 = 3 # ^60 - h h 20 = 60 - h h = 40 m Hence, Height of tower is 40 m. 9. The angle of elevation of the top of a building from the foot of the tower is 30c and the angle of elevation of the top of the tower from the foot of the building is 60c. If the tower is 60 m high, find the height of the building. Ans : [CBSE Delhi 2013] Here A is cloud and Al is refection of cloud. In right TAOP we have tan 30c = H - 120 OP Let AB be the tower of 60 m height and CD be the building of h height. As per given in question we have drawn figure below. 1 = H - 120 OP 3 OP = ^H - 120h 3 ...(1) In right TOPA' we have tan 60c = H + 120 OP OP = H + 120 3 ...(2) From (1) and (2), we get H + 120 = 3 H - 120 ^ h 3 Thus height of cloud is 240 m. 11. As observed from the top of a light house, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, changes from 30c to 60c . Find the distance travelled by the ship during the Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 189 Chap 9 : Some Applications of Trigonometry www.cbse.online period of observation. (Use 3 = 1.73 ) Ans : [CBSE Outside DelhiI 2016] Let AB be the light house of height 100 m. Let C and D be the position of ship at elevation 60c and 30c. As per given in question we have drawn figure below. Let the speed of car be x m/sec. Thus distance covered in 6 sec = 6x . Hence DC = 6x m Let distance (remaining) CA covered in t sec. CA = tx Now in right TADB , In right TABC we have AB = tan 60 BC 100 = y 3 y = 100 3 In right TABD, we have AB = tan 30 BD 100 = 1 x 3 x = 100 3 Required distance travelled by ship, x - y = 100 3 - 100 m 3 = 100 ; 3 - 1E 3 100 2 # = 3 = 200 = 200 3 3 3 CD = x - y = 200 # 1.73 = 3.46 m 3 3 AD = AC + CD = 6x + 9x h tan 30c = 6x + tx h = 6+t x 3 ...(1) In right TACB we have, tan 60c = h tx 3t = h tx ...(2) From eqn. (1) and (2) we get 3t = 6+t 3 3t = 6 + t 2t = 6 t =3 Hence, car takes 3 seconds. 13. An angle of elevation of a cloud from a point 60m above the surface of the water of a lake is 30c and the angle of depression of its shadow in water is 60c. Find the height of the cloud from the surface of water. Ans : [CBSE Delhi Set-I] As per given in question we have drawn figure below. 115.33 m 12. A straight highway leads to the foot of a tower. A man standing on its top observes a car at an angle of depression of 30c, which is approaching the foot of the tower with a uniform speed. 6 seconds later, the angle of depression of the car becomes 60c. Find the time taken by the car to reach the foot of tower from this point. Ans : [Delhi Compt Set-I, III, 2017] Let AB be the tower of fight h . Let point C and D be location of car. As per given in question we have drawn figure below. Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 190 Chap 9 : Some Applications of Trigonometry Here h = tan 30c = 1 x 3 x =h 3 and www.rava.org.in h 3 - h = 50 3 ...(1) h + 60 + 60 = tan 60c x h + 120 = x 3 h + 120 = x 3 ...(2) From (1) and (2) we get h + 120 = 3h# h ^ 3 - 1h = 50 3 50 3 ^ 3 - 1h h = 50 3 = 3 -1 ^ 3 - 1h^ 3 + 1h 50 ^3 - 3 h = 3-1 h = 25 ^3 + 3h = 25 # 4.732 = 118.3 m Hence, the height of tower = distance between 3 building and tower = 118.3 m h + 120 = 3h h = 120 = 60 m 2 Hence height of cloud from surface of water 15. An observer finals the angle of elevation of the top of = 60 + 60 = 120 m 14. The angle of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30c and 45c respectively. Find the height of the tower and also the horizontal distance between the building and the tower. Ans : [Sample Question Paper 2017-18] the tower from a certain point on the ground as 30c . If the observer moves 20 m. Towards the base of the tower, the angle of elevation of the top increase by 15c , find the height of the tower. Ans : [CBSE Delhi Set-III 2017] Let AB be the tower of height h . Angle of elevation from point D and C are given 30c and 45c respectively. As per given in question we have drawn figure below. Let CD be the building of height 50 m and AB be the tower of height h . Angle of depressions of top and bottom are given 30c and 60c respectively. As per given in question we have drawn figure below. Here CB = x and DC = 20 m Now in right TABC , AB = tan 45c BC h =1 x h =x In right TABC we have AB = tan 30c DB Let distance between BO be x . Now, in right TABD AB = tan 45c BD h = 1 ^20 + x h 3 h 3 = 20 + x h = 10 x h =x ...(1) h 3 = 20 + h In right TAEC we have AE = tan 30c , EC h 3 - h = 20 h ^ 3 - 1h = 20 h - 50 = 1 x 3 x = h 3 - 50 3 From (1) and (2) we get h = h 3 - 50 3 Substituting the value of x from (1) in (2) 20 ^ 3 + 1h 20 = 3 -1 ^ 3 - 1h^ 3 + 1h 20 ^ 3 + 1h = 3-1 h = ...(2) = 10 ^ 3 + 1h Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 191 Chap 9 : Some Applications of Trigonometry www.cbse.online Hence, the height of tower = 10 ^ 3 + 1h m 16. From a point P on the ground, the angles of elevation of the top of a 10 m tall building and a helicopter, hovering at some height vertically over the top of the building are 30c and 60c respectively. Find the height of the helicopter above the ground. Ans : [CBSE Outside Delhi Compt. 2017] Let AB be the building of height 10 m and the height of the helicopter from top the building be x . As per given in question we have drawn figure below. In right TDCA we have DC = tan 60c CA 15 = x 3 x = 15 = 5 3 3 In right TDCB we have DC = tan 45c CB 15 z = 1 x+y x + y = 15 Let the distance between point and building be y . Height of the helicopter from ground 5 3 + y = 15 = ^10 + x h m y = 15 - 5 3 = 5 ^3 - In right TBAP we have AB = tan 30c BP Hence, the distance between points = 5 ^3 - 10 = 1 y 3 ...(1) y = 10 3 In right TCAP , AC = tan 60c PA 10 + x = y 3h m 3h m 18. From the top of a tower, 100 m high, a man observes two cars on the opposite sides of the tower and in same straight line with its base, with angles of depression 30c and 45c. Find the distance between the cars. [ Take 3 = 1.732 ] Ans : [CBSE Outside Delhi Compt. Set-III 2017] Let BD be the tower of height 100 m. Let A and C be location of car on opposite side of tower As per given in question we have drawn figure below. 3 10 + x = y 3 ...(2) From (1) and (2) 10 + x = 10 3 # 3 = 40 x = 30 Hence height of the helicopter is 30 m. Add 8905629969 in Your Class Whatsapp Group to Get All PDFs 17. Two points A and B are on the same side of a tower and in the same straight line with its base. The angle of depression of these points from the top of the tower are 60c and 45c respectively. If the height of the tower is 15 m, then find the distance between these points. Ans : [CBSE Delhi 2017] Let CD be the tower of height 15 m. Let A and B point on same side of tower As per given in question we have drawn figure below. In right TABD, +DAB = 30c In TBDC , +BCD = 45c also, BD = 100 m In right TABD we have, tan 30c = DB AB Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 192 Chap 9 : Some Applications of Trigonometry www.rava.org.in 20. An aeroplane is flying at a height of 300 m above the 1 = 100 AB 3 ground. Flying at this height the angle of depression from the aeroplane of two points on both banks of a respectively. Find the width of the river. River in opposite direction are 45c and 60c. Ans : [CBSE Outside Delhi Set-I 2017] AB = 100 3 m In right TDBC we have, tan 45c = DB BC Let A be helicopter flying at a height of 300 m above the point O on ground. Let B and C be the bank of river. As per given in question we have drawn figure below. 1 = 100 BC BC = 100 m Now, AB + BC = 100 + 100 3 = 100 ^ 3 + 1h = 100 + 173.2 = 273.2 m 19. The angle of depression of two ships from an aeroplane flying at the height of 7500 m are 30c and 45c. if both the ships are in the same that one ship is exactly behind the other, find the distance between the ships. Ans : [CBSE Foreign 2017] Let A, C and D be the position of aeroplane and two ship respectively. Aeroplane is flying at 7500 m height from point B . As per given in question we have drawn figure below. Let BO be x and OC be y . In right TAOC we have AO = tan 45c OC 300 = 1 y y = 300 In right TAOB we have AO = tan 60c BO 300 = x 3 x 3 = 300 x = 300 = 100 3 3 BC = y + x = 300 + 100 3 In right TABC we have AB = tan 45c BC = 300 + 100 # 1.732 = 473.2 m Hence, the width of river is 473.2 m 7500 = y y y = 7500 21. From the top of a hill, the angle of depression of two ...(1) In right TABD we have AB = tan 30c BD Let AB be the hill of height h . Angle of depression from point D and C are given 30c and 45c respectively. As per given in question we have drawn figure below. 7500 = 1 x+y 3 x + y = 7500 3 consecutive kilometre stones due east are found to be 45c and 30c respectively. Find the height of the hill. [Use 3 = 1.73 ] Ans : [CBSE Outside Delhi 2016] ...(2) Substituting the value of y from (1) in (2) we have x + 7500 = 7500 3 x = 7500 3 - 7500 = 7500 ^ 3 - 1h = 7500 ^1.73 - 1h = 7500 # 0.73 = 5475 m Hence, the distance between two ships is 5475 m. Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 193 Chap 9 : Some Applications of Trigonometry www.cbse.online x =1 y 3 In right TABC we have AB = tan 45c AC h =1 x Thus x :y = 1:3 23. From the top of a 7 m high building, the angle of elevation of the top of a tower is 60c and the angle of depression of its foot is 45c. Find the height of the tower. (Use 3 = 1.732 ) Ans : [Foreign Set-I, II] h =x In right TABD we have AB = tan 30c AC + CD Let AB be the building of height 7 m and CD be the tower of height h . Angle of depressions of top and bottom are given 30c and 60c respectively. As per given in question we have drawn figure below. h = 1 x + 1000 3 h 3 = h + 1000 h ^ 3 - 1h = 1000 1000 ^ 3 + 1h h = 1000 = 3 -1 ^ 3 - 1h^ 3 + 1h 1000 ^ 3 + 1h = 3-1 = 500 ^ 3 + 1h = 500 ^1.73 + 1h = 500 # 2.73 = 1365 Hence height of the hill is 1365 m. 22. The tops of two towers of height x and y , standing on level ground, subtend angles of 30c and 60c respectively at the centre of the line joining their feet, then find x : y . Ans : [Delhi CBSE Term-2, 2015] Let AB be the tower of height x and CD be the tower of height y . Angle of depressions of both tower at centre point M are given 30c and 60c respectively. As per given in question we have drawn figure below. Here +CBD = +ECB = 45c due to alternate angles. In right TABC we have CD = tan 45c BD 7 =1 x x =7 In right TAEC we have CE = tan 60c AE h-7 = x Here M is the centre of the line joining their feet. Let BM = MD = z h-7 = x 3 h-7 = 7 3 In right TABM we have, x = tan 30c z h = 7 3 +7 = 7 ^ 3 + 1h = 7 ^1.732 + 1h x = z# 1 3 Hence, height of tower = 19.124 m In right TCDM we have y = tan 60c z y = z# From (1) and (2), we get 3 1 x = z# 3 y z# 3 3 HOTS QUESTIONS 1. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun s altitude is 30c, than when it is 60c. Find the height of the tower. Ans : [CBSE Board Term-2, 2011] Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 194 Chap 9 : Some Applications of Trigonometry www.rava.org.in Let AB be the tower of height h . Let BC be the shadow at 60c and BD be shadow at 30c. As per given in question we have drawn figure below. In right TABD we have, tan 45c = AB = 1 BD 1 = 50 x x = 50 m (1) Thus distance of pole from bottom of tower is 50 m. Now in TAMC we have tan 30c = AM = AM x MC AM = 50 or 28.87 m. 3 (2) Height pole = 50 - 28.87 = 21.13 m. In right TABC we get, tan 60c = AB BC 3. 3 =h x h = 3x In right TABD we have, tan 30c = 3h = The angle of elevation of an aeroplane from a point A on the ground is 60c. After a flight of 15 seconds, the angle of elevation changed to 30c. If the aeroplane is flying at a constant height of 1500 3 m, find the speed of the plane in km/hr. Ans : [CBSE Outside Delhi 2015] Let A be the point on ground, B and C be the point of location of aeroplane at height of 1500 3 m. As per given in question we have drawn figure below. AB BC + 40 1 = h x + 40 3 x + 40 = h = CD = BM 3# 3 x = 3x 40 = 2x & x = 20 m h = 3 # 20 = 20 3 m Thus height of tower is 20 3 m For more files visit www.cbse.online 2. From the top of a tower of height 50 m, the angles of depression of the top and bottom of a pole are 30c and 45c respectively. Find : (1) How far the pole is from the bottom of the tower, (2) The height of the pole. (Use 3 = 1.732 ) Ans : [CBSE Foreign 2015] Let AB be the tower of height 50 m and CD be the pole of height h . From the top of a tower of height 50 m, the angles of depression of the top and bottom of a pole are 30c and 45c respectively. As per given in question we have drawn figure below. In right TBAL BL = tan 60c AL 1500 3 = x 3 BL = CM x = 1500 m. In right TCAM we have CM = tan 30c AL + LM 1500 3 = 1 x+y 3 x + y = 1500 # 3 1500 + y = 4500 y = 3000 m. y Speed = Dis tan ce = t Time = 3000 = 200 m/s 15 = 200 # 60 # 60 1000 Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 195 Chap 9 : Some Applications of Trigonometry www.cbse.online on the opposite bank is from the bank, he finds 30c. Find the height of the river. Ans : = 720 km/hr. Hence, the speed of the aeroplane is 720 km/hr. For more files visit www.cbse.online 4. At a point A, 20 metre above the level of water in a lake, the angle of elevation of a cloud is 30c. The angle of depression of the reflection of the cloud in the lake, at A is 60c. Find the distance of the cloud from A ? Ans : [CBSE Outside Delhi, 2015] 60c. When he retreats 20 m the angle of elevation to be the tree and the breadth of [CBSE Board Term-2, 2012] Let AB be the tree of height h and breadth of river be b . As per given in question we have drawn figure below. Here point C and D are the location of person . As per given in question we have drawn figure below. Here cloud is at C , D is reflection of cloud in water. In right TABC we have, h = tan 60c = b h = ...(1) x = 3h Here DE = EC because D is reflection of cloud and E is at water level. EC + EB = x 3 h + 20 + 20 = x 3 h + 40 = 3b = b + 20 & b = 10 m h = b 3 = 10 # 1.73 = 17.3 m Thus height of tree is 17.3 m and breadth of river is 10 m. 6. ...(2) 3x From (1) and (2), h + 40 = 3# h = 20 m 3 h = 3h x = 20 3 Now AC = = = ...(2) From (1) and (2) we have b 3 = b + 20 3 In right TABD we have BD = tan 60c BA 3 ...(1) 3b In right TABD we have h = tan 30c = 1 b + 20 3 b 20 + h = 3 In right TABC we have h = tan 30c = 1 x 3 DC + EB = x 3 A boy observes that the angle of elevation of a bird flying at a distance of 100 m is 30c. At the same distance from the boy, a girl finds the angle of elevation of the same bird from a building 20 m high is 45c. Find the distance of the bird from the girl. Ans : [CBSE Borad Term-2, 2014] Let O be the position of the bird and B be the position of the boy. Let FG be the building and G be the position of the girl. As per given in question we have drawn figure below. 2 2 ^BC h + ^AB h ^20h2 + ^20 3 h 400 + 1200 2 = 40 m. Hence distance of the cloud is 40 m. 5. A person standing on the bank of a river, observes that the angle of elevation of the top of the tree standing Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 196 Chap 9 : Some Applications of Trigonometry www.rava.org.in tan 45c = 80 y y = 80 In right DOC we have tan 30c = 80 x+y 1 = 80 x+y 3 x + y = 80 3 x = 80 3 - y = 80 3 - 80 In right TOLB we have OL = sin 30c BO OL = 1 2 100 OL = 50 m OM = OL - ML = OL - FG = 50 - 20 = 30 m In right TOMG we have OM = sin 45c OG x = 80 ^ 3 - 1h = 58.4 m. Hence, speed of bird = 58.4 = 29.2 m 2 8. The angle of elevation of a cloud from a point 200 m above the lake is 30c and the angle of depression of its reflection in the lake is 60c, find the height of the cloud above the lake. Ans : [CBSE Board Term-2 2012 Set (59), 2011, Set B1] Let h be the height of cloud at C from lake. Let x be the horizontal distance of cloud from point of A. As per given in question we have drawn figure below. OM = 1 OG 2 30 = 1 OG 2 OG = 30 2 m Hence, distance of the bird from the girl is 30 2 m. For more files visit www.cbse.online 7. A bird sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45c. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30c. Find the speed of flying of the bird. (Take 3 = 1.732 ) Ans : [CBSE Delhi Set I, III, 2016] Let CD be the tree of height 80 m and bird is sitting at D . Point O on ground is reference point from where we observe bird. As per given in question we have drawn figure below. Here BE is water level of lake and F is the reflection of cloud seen from A. In right TADC we have tan 30c = h - 200 x 1 = h - 200 x 3 x = 3 ^h - 200h ...(1) In right TADF we have tan 60c = h + 200 x 3 = h + 200 x 3 x = h + 200 ...(2) From (1) and (2) we have 3 ^h - 200h = h + 200 3h - h = 200 + 600 2h = 800 In right AOB we have So, height of cloud H = 400 m. Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 197 Chap 9 : Some Applications of Trigonometry 9. www.cbse.online The angle of elevation of a jet fighter point A on ground is 60c. After flying 19 seconds, the angle changes to 30c. If the jet is flying at a speed of 648 km/hour, find the constant height at which the jet is flying. Ans : [CBSE Board Term-2, 2012] Let C and D are the point of location of jet at height h . Point B and E are foot print on ground of get at thee location. As per given in question we have drawn figure below. Here distance covered in 2 minutes is 2x . Thus CD = 2x In right TABD we have AB = tan 60c BC 150 = y 3 y = 150 = 50 3 3 In right TABD we have AB = tan 45c BD In 3600 sec distance travelled by plane = 648000 m In 10 sec distance travelled by plane - 648000 # 10 3600 = 1800 m In right TABC, we have h = tan 60 = x 150 = 1 y + 2x y + 2x = 150 3 h =x 3 In right TADE we have h = tan 30 = 1 x + 1800 3 x 1800 + h = 3 ...(1) ...(2) From equations (1) and (2), we get ...(1) 50 3 + 2x = 150 2x = 150 - 50 3 ...(2) = 1558.5 m Thus height of jet is 1558.8 m. 10. A moving boat observed from the top of a 150 m high cliff. moving away from the cliff. The angle of depression of the boat changes from 60c to 45c in 2 minutes. Find the speed of the boat. Ans : [Delhi Set-I 2017] 3h 3 h m/min. 25 ^3 - 3 h # 60 1000 = 3 ^3 2 3 h km/hr. For more files visit www.cbse.online 2x = 1800 = 900 # 1.732 x = 25 ^3 - = 3x = x + 1800 h =x 3 3h Speed of the boat = 25 ^3 - From equations (1) and (2), we get x 3 = x + 1800 3 x = 900 m 2x = 50 ^3 - 11. From the top of a 7 m high building the angle of elevation of the top of a tower is 60c and the angle of depression of its foot is 45c. Find the height of the tower. Ans : [CBSE Delhi Set-I, II, III, 2017] Let AB be the building of height 7 m and CD be the tower of height h . Let distance between two be x . Angle of depressions of top and bottom of tower are given 60c and 45c respectively. As per given in question we have drawn figure below. Let AB be building = 7 m Let AB be the cliff of height 150 m. Let C and D be the point of boat at 60c and 45c. Let the speed of the boat be x m/min. Let BC be y As per given in question we have drawn figure below. Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 198 Chap 9 : Some Applications of Trigonometry www.rava.org.in AB = tan 45c AD 120 = 1 AB AB = 120 In right TBAC we have AB = tan 60c CA 120 = CA 3 CA = 120 = 40 3 3 CD = CA + AD = 120 + 40 3 = 120 + 40 # 1.732 = 189.28 m CD be the height of tower = ^7 + h h m Hence the distance between two men is 189.28 m. BD = AE = x m For more files visit www.cbse.online In right TABD we have AB = tan 45c BD NO NEED TO PURCHASE ANY BOOKS 7 =1 x x =7 m ...(1) In right TCEA we have CE = tan 60c AE h-7 = x 3 h-7 = x 3 ...(2) For session 2019-2020 free pdf will be available at www.cbse.online for 1. Previous 15 Years Exams Chapter-wise Question Bank 2. Previous Ten Years Exam Paper (Paper-wise). 3. 20 Model Paper (All Solved). 4. NCERT Solutions All material will be solved and free pdf. It will be provided by 30 September and will be updated regularly. Disclaimer : www.cbse.online is not affiliated to Central Board of Secondary Education, New Delhi in any manner. www.cbse.online is a private organization which provide free study material pdfs to students. At www.cbse.online CBSE stands for Canny Books For School Education From equations (1) and (2), we get h-7 = 7 3 h = 7 + 7 3 = 7 ^1 + Hence, the height of tower is 7 ^1 + 3h m 3h m 12. From the top of a 120 m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of repression as 60c and 45c. Find the distance between two cars. Ans : [Delhi Compt. Set-III, II, 2017] Let AB be the tower of height 120 m. Let C and D be location of car on opposite side of tower. As per given in question we have drawn figure below. In right TBAD we have Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 199

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

 

  Print intermediate debugging step

Show debugging info


 

 

© 2010 - 2025 ResPaper. Terms of ServiceContact Us Advertise with us

 

sumayya35 chat