Question Bank Trigonometry

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1 Question Bank Trigonometry cos A sin A cos A sin A 1. Prove that cos A sina cos A sina cos A sin A cos A sin A L.H.S. cos A sina cos A sina (cosa sina) (cos A sin A cosa sina) (cosa sina) cos A sin A cosa sina (cosa sina) (cosa sina) a b ( a b) ( a b ab) and a b ( a b) ( a b ab) (cos A sin A cosasina) (cos A sin A cosa sina) (1 cosa sina) (1 cosa sina) cos A sin A 1 1 cosa sina 1 cosa sina R.H.S. Proved. cosa sina. Prove that cosa sina 1 tana 1 cota cosa sina L.H.S. 1 tana 1 cota cosa sina sina cosa 1 1 cosa sina cosa cosa cosa sina sina sina cosa sina sin A cosa sin A cos A cosa sina cos A sin A cosa sina (cosa sina) (cosa sina) (cosa sina) Math Class X 1 Question Bank

2 [ a b (a b) (a b)] cosa sina R.H.S. Proved. sina 3. Prove that cota coseca sina L.H.S. cota coseca sina cota coseca sina sina sina cosa 1 cosa 1 cosa 1 sina sina sin A 1 cos A sin 1 cos cosa 1 θ (1 cosa)(1 cosa) 1 cosa (cosa 1) sina R.H.S. cota coseca sina sin A cosa 1 cosa 1 sina sina 1 cos A sin 1 cos cosa 1 θ (1 cosa) (1 cosa) cosa 1 (1 cosa) (1 cosa) cosa 1 θ θ (1 cosa) 1 cos A L.H.S. Proved. Math Class X Question Bank

3 4. If sina cosa m and seca coseca n, prove that n (m 1) m. We have, m sina cosa m (sina cosa) sin A cos A sina cosa 1 sin A cosa m 1 1 sina cosa 1 sina cosa n (m 1) (seca coseca). sina cosa sina cosa seca sina cosa coseca sina cosa [ cosa seca 1 and sina cosec A 1] (sina cosa) m Hence, n (m 1) m. Proved. 5. Prove that (sec A tana) cosa cosa (seca tana) 1 1 L.H.S. seca tana cosa sina cosa cos A cosa cosa 1 cos A 1 sina 1 sina cosa cos A ( 1 sina) 1 sin A 1 sina sina(1 sina) cos A(1 sina) tana. cosa (1 sina) Math Class X 3 Question Bank

4 1 1 R.H.S. cosa seca tana 1 1 cosa 1 sina cosa cosa 1 cosa cosa 1 sina 1 sina cos A cosa (1 sina) 1 sina 1 sin A sina(1 sina) cos A(1 sina) cos A(1 sina) tana. Hence, LHS RHS. Proved. 6. If x sin 3 θ y cos 3 θ sinθ and x sinθ y 0, then prove that x y 1 We have x sin 3 θ y cos 3 θ sinθ... (i) x sinθ y 0... (ii) sin θ y x (iii) From (i) sin θ cos θ x. y. 1 sinθ sinθ x. sin θ y. cos θ 1 sinθ x. y x sin θ y.. 1 x y [From (iii)] y sinθ x 1 x y sinθ 1...(iv) Squaring (ii) and (iv) and adding, we get, (x sinθ y ) (x y sinθ) 0 1 x sin θ y cos θ xy sinθ x cos θ y i sin θ xy sinθ 1 Math Class X 4 Question Bank

5 x ( sin θ cos θ) y (cos θ sin θ) 1 x y 1. Proved. 7. Is an identity? If not solve for θ, cosec θ 1 cosec θ 1 where 0 < θ < 90. Here, LHS cosec θ 1 cosec θ sin θ sinθ cos θ sinθ cos θ sinθ 1 sinθ 1 sinθ cos θ sin θ ( 1 sin θ 1 sin θ) 1 sin θ sin θ tanθ cos θ Thus, the given equality becomes tanθ If the equality holds true for all values of θ, then the equality is an identity. Let us take θ 30 So, tan θ tan30 3 tan θ for θ 30 Therefore the equality is not an identity. It is an equation. Now, tan θ tan θ 1 tan θ tan 45 θ If tan θ sec θ 3, where θ is acute, then prove that 5 sinθ 4. We have tanθ secθ 3 sinθ sinθ 3 Math Class X 5 Question Bank

6 (1 sinθ) 9 cos θ [Squaring both sides] 1 sin θ sinθ 9 9 sin θ 10 sin θ sinθ sin θ 10 sinθ 8 sinθ sinθ (sinθ 1) 8 (sinθ 1) 0 (sinθ 1) (10 sinθ 8) 0 sinθ 1 or sinθ sin θ 4 [Rejecting sinθ 1, since θ is acute] 5 5 sinθ 4. Proved. 9. Without using trigonometric tables, prove that : tan 10º tan 0º tan 1 30º tan 70º tan 80º 3 L.H.S. tan 10º tan 0º tan 30º tan 70º tan 80º (tan 10º tan 80º ), (tan 0º tan 70º) tan 30º tan (90º 80º) tan 80º. tan (90º 70º) tan 70º tan 30º cot 80º tan 80º. cot70º tan 70º tan 30º [ tan (90º θ ) cot θ ] 1 1. tan80º. tan 70º tan 30º tan 80º tan 70º 1.1. tan 30º 1 R.H.S. Proved Prove that sina cosa sec (90º A ) cosec (90º A) sin(90º A) cos(90º A) sina cosa L.H.S. sin(90º A) (cos(90º A) sina cosa cosa sina Math Class X 6 Question Bank

7 [ sin (90º A) cosa and cos (90º A) sina] sin A cos A 1 sin A cos A 1 sin A cos A sina cosa coseca seca R.H.S sec ( 90º A ) cosec (90º A) coseca seca [ sec (90º A) cosec A and cosec (90º A) seca] L.H.S. Hence, L.H.S. R.H.S. Proved. 11. Prove that L.H.S. sin 0º sin 70º sin (90º θ) sinθ cos(90º θ) cos 0º cos 70º tanθ cotθ sin 0º sin 70º sin (90º θ) sin θ cos(90º θ) cos 0º cos 70º tan θ cotθ sin (90º 70º ) sin 70º cos sin sin cos θ θ θ θ cos (90º 70º) cos 70º tan θ cotθ cos 70º sin 70º cos θ sinθ sin θ sin 70 cos 70º sinθ cos θ sin θ [ sin (90º θ) cos θ,cos (90º θ ) sinθ] 1 cos θ sin θ R.H.S. Proved. 1. Using the tables, find the values of (i) sin 60º 3 (ii) cos 1º 56 (iii) tan 75º (iv) cot 40º 36 From trigonometric tables, we have (i) sin 60º Mean difference for 5 7 (To be added) sin 60º (ii) cos 51º Mean difference for (To be subtracted) Math Class X 7 Question Bank

8 1º (iii) tan 75º Mean difference for 93 (To be added) 75º (iv) cot 40º 36 cot (90º 49º 4 ) tan 49º 4 Now, tan 49º Find θ when (i) sin θ (ii) cos θ (iii) tan θ (i) From the table, find the angle whose sine is just smaller than sin θ sin 5º Difference Mean difference 14 corresponds to 5 Required angle (5º º 41. (ii) From the table, find the angle whose cosine is just greater than cos θ sin 5º Difference Mean difference 1 corresponds to 5 Required angle (56º 18 5 ) 56º 3. (iii) From the table, find the angle whose tangent is just smaller than tan θ tan 79º Since mean differences are not given corresponding to 79, therefore required angle Math Class X 8 Question Bank

9 14. A boy standing on a vertical cliff in a jungle observes two rest houses in line with him on opposite sides deep in the jungle below. If their angles of depression are 19 and 6 and the distance between them is m, find the height of the cliff. Let A be the top of the cliff and C and D be the two rest houses. Let AB h m and BC x m Then, BD ( x) m In ΔABC, tan 19 h x h x h x (i) h In ΔABD, tan 6 x h x h ( x) (ii) From (i) and (ii), we have ( x) x x ( ) x 0.83 From (i), we have, h log h log log log log ( 1 1 1) ( ) h antilog Hence, height of the cliff m. Math Class X 9 Question Bank

10 13. An aeroplane is flying horizontally 4000 m above the ground and is going away from an observer on the level ground. At a certain instant the observer finds that the angle of elevation of the plane is 45. After 15 seconds, its elevation from the same point changes to 30. Find the speed of the aeroplane in km/h. Let A be the position of the observer, B be the point whose angle of elevation from A is 45. Let after 15 seconds the position of the plane be C, whose angle of elevation from A be 30. In ΔABD, tan 45 BD AD 1 BD AD BD AD..(i) CE 4000 In ΔACE, tan 30 AE AD DE [ AE AD DE] BD DE [From (i)] DE DE DE 4000( 3 1 ) DE Distance covered by the aeroplane in 15 seconds 98 m Speed of the aeroplane m/s km/h 70.7 km/h. Math Class X 10 Question Bank

11 14. At the foot of a mountain, the elevation of its summit is 45. After ascending 1000 m towards the mountain up a slope of 30 inclination, the elevation is found to be 60. Find the height of the mountain. Let AB be the mountain of height h m and C be its foot. CD 1000 m, ACB 45, DCB 30 and ADF 60. h In ΔACB, tan 45 CB h 1 CB h CB..(i) In ΔCDE, sin 30 DE DE 1000 DE 500..(ii) In ΔCDE, cos 30 CE CE 1000 CE (iii) Now, BE BC EC h [From (i) and (iii)] In Δ ADF, tan 60 AF DF h BE h h h h 500 h h ( 3 1 ) 1000 h [ DF BF and BF DE] Math Class X 11 Question Bank

12 Hence, height of the mountain is m. 15. A man is standing on the deck of a ship which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60 and the angle of depression of the base of the hill as 30. Calculate the distance of the hill from the ship and the height of the hill. In the figure. A is the deck of the ship and CD is the hill. Let BC x m and DE h m. In ΔABC, tan 30 8 x x x 8 3m. In ΔADE, tan 60 DE AE h x 3 h x [AE BC x] h 3 x cm. Distance of the hill from the ship 8 3m, and height of the hill (h 8) m (4 8) m 3 m. 16. A ladder rests against a house on one side of a street. The angle of elevation of the top of the ladder is 60. The ladder is turned over to rest against a house on the other side of the street and the elevation now becomes 4 50'. If the ladder is 40 m long, find the breadth of the street. In the figure, AB and CD are two houses. O is a point on the street, at which one end of the ladder rests. Let OB x m and OD y m. x In ΔAOB, cos x x 0 m. 40 Math Class X 1 Question Bank

13 In ΔCOD, cos 4 50' 40 y y [From tables] y Hence, breadth of the street (x y) m (0 9.33) m m. 17. A vertical tower stands on horizontal plane and is surmounted by a vertical flagstaff of height h m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is α and that of the top of the h tanα flagstaff is β. Prove that the height of the tower is tanβ tanα. Let AB be the tower, AC be the flagstaff of height h m and D be the point of observation. AB BD In ΔABD, tan α AB BD tan α..(i) In ΔCBD, tan β BC BD AB AC tan β BD AB AC BD tanβ BD tan α h BD tanβ h BD (tanβ tanα) h BD tanβ tan α Height of the tower AB BD tan α h tanα tanβ tanα [From (i)] Math Class X 13 Question Bank