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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10

Samacheer Kalvi solutions for Mathematics [English] Class 10 SSLC TN Board chapter 6 - Trigonometry [Latest edition]

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Samacheer Kalvi solutions for Mathematics [English] Class 10 SSLC TN Board chapter 6 - Trigonometry - Shaalaa.com
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Solutions for Chapter 6: Trigonometry

Below listed, you can find solutions for Chapter 6 of Tamil Nadu Board of Secondary Education Samacheer Kalvi for Mathematics [English] Class 10 SSLC TN Board.


Exercise 6.1Exercise 6.2Exercise 6.3Exercise 6.4Exercise 6.5Unit Exercise – 6
Exercise 6.1 [Pages 249 - 250]

Samacheer Kalvi solutions for Mathematics [English] Class 10 SSLC TN Board 6 Trigonometry Exercise 6.1 [Pages 249 - 250]

Exercise 6.1 | Q 1. (i) | Page 249

Prove the following identities.

cot θ + tan θ = sec θ cosec θ

Exercise 6.1 | Q 1. (ii) | Page 249

Prove the following identities.

tan4 θ + tan2 θ = sec4 θ – sec2 θ

Exercise 6.1 | Q 2. (i) | Page 249

Prove the following identities.

`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ

Exercise 6.1 | Q 2. (ii) | Page 249

Prove the following identities.

`costheta/(1 + sintheta)` = sec θ – tan θ

Exercise 6.1 | Q 3. (i) | Page 250

Prove the following identities.

`sqrt((1 + sin theta)/(1 - sin theta)` = sec θ + tan θ

Exercise 6.1 | Q 3. (ii) | Page 250

Prove the following identities.

`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ

Exercise 6.1 | Q 4. (i) | Page 250

Prove the following identities.

sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1

Exercise 6.1 | Q 4. (ii) | Page 250

Prove the following identities.

(sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2

Exercise 6.1 | Q 5. (i) | Page 250

Prove the following identities.

sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1

Exercise 6.1 | Q 5. (ii) | Page 250

Prove the following identities.

`(cot theta - cos theta)/(cot theta + cos theta) = ("cosec"  theta - 1)/("cosec"  theta + 1)`

Exercise 6.1 | Q 6. (i) | Page 250

Prove the following identities.

`(sin "A" - sin "B")/(cos "A" + cos "B") + (cos "A" - cos "B")/(sin "A" + sin "B")`

Exercise 6.1 | Q 6. (ii) | Page 250

Prove the following identities.

`(sin^3"A" + cos^3"A")/(sin"A" + cos"A") + (sin^3"A" - cos^3"A")/(sin"A" - cos"A")` = 2

Exercise 6.1 | Q 7. (i) | Page 250

If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1

Exercise 6.1 | Q 7. (ii) | Page 250

If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`

Exercise 6.1 | Q 8. (i) | Page 250

If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2

Exercise 6.1 | Q 8. (ii) | Page 250

If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1

Exercise 6.1 | Q 9. (i) | Page 250

If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.

Exercise 6.1 | Q 9. (ii) | Page 250

If sin θ (1 + sin2 θ) = cos2 θ, then prove that cos6 θ – 4 cos4 θ + 8 cos2 θ = 4

Exercise 6.1 | Q 10 | Page 250

If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ

Exercise 6.2 [Page 257]

Samacheer Kalvi solutions for Mathematics [English] Class 10 SSLC TN Board 6 Trigonometry Exercise 6.2 [Page 257]

Exercise 6.2 | Q 1 | Page 257

Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height `10sqrt(3)` m

Exercise 6.2 | Q 2 | Page 257

A road is flanked on either side by continuous rows of houses of height `4sqrt(3)` m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30°. Find the width of the road

Exercise 6.2 | Q 3 | Page 257

To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? `(sqrt(3) = 1.732)`

Exercise 6.2 | Q 4 | Page 257

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 60ϒ and from the same point the angle of elevation of the top of the pedestal is 40ϒ. Find the height of the pedestal. (tan 40° = 0.8391, `sqrt(3)` = 1.732)

Exercise 6.2 | Q 5. (i) | Page 257

A flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30°. Find the height of the pole `(sqrt(3) = 1.732)`

Exercise 6.2 | Q 5. (ii) | Page 257

A flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30°. Find radius of the dome `(sqrt(3) = 1.732)`

Exercise 6.2 | Q 6 | Page 257

The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?

Exercise 6.3 [Page 261]

Samacheer Kalvi solutions for Mathematics [English] Class 10 SSLC TN Board 6 Trigonometry Exercise 6.3 [Page 261]

Exercise 6.3 | Q 1 | Page 261

From the top of a rock `50sqrt(3)` m high, the angle of depression of a car on the ground is observed to be 30°. Find the distance of the car from the rock

Exercise 6.3 | Q 2 | Page 261

The horizontal distance between two buildings is 70 m. The angle of depression of the top of the first building when seen from the top of the second building is 45°. If the height of the second building is 120 m, find the height of the first building

Exercise 6.3 | Q 3 | Page 261

From the top of the tower 60 m high the angles of depression of the top and bottom of a vertical lamp post are observed to be 38° and 60° respectively. Find the height of the lamp post (tan 38° = 0.7813, `sqrt(3)` = 1.732)

Exercise 6.3 | Q 4 | Page 261

An aeroplane at an altitude of 1800 m finds that two boats are sailing towards it in the same direction. The angles of depression of the boats as observed from the aeroplane are 60° and 30° respectively. Find the distance between the two boats. `(sqrt(3) = 1.732)`

Exercise 6.3 | Q 5 | Page 261

From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it is observed to be 30° and 60°. If the height of the lighthouse is h meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is `(4"h")/sqrt(3)` m

Exercise 6.3 | Q 6 | Page 261

A lift in a building of height 90 feet with transparent glass walls is descending from the top of the building. At the top of the building, the angle of depression to a fountain in the garden is 60°. Two minutes later, the angle of depression reduces to 30°. If the fountain is `30sqrt(3)` feet from the entrance of the lift, find the speed of the lift which is descending.

Exercise 6.4 [Pages 264 - 265]

Samacheer Kalvi solutions for Mathematics [English] Class 10 SSLC TN Board 6 Trigonometry Exercise 6.4 [Pages 264 - 265]

Exercise 6.4 | Q 1 | Page 264

From the top of a tree of height 13 m the angle of elevation and depression of the top and bottom of another tree are 45° and 30° respectively. Find the height of the second tree. `(sqrt(3) = 1.732)`

Exercise 6.4 | Q 2 | Page 265

A man is standing on the deck of a ship, which is 40 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 `(sqrt(3) = 1.732)`

Exercise 6.4 | Q 3 | Page 265

If the angle of elevation of a cloud from a point ‘h’ metres above a lake is θ1 and the angle of depression of its reflection in the lake is θ2. Prove that the height that the cloud is located from the ground is `("h"(tan theta_1  +  tan theta_2))/(tan theta_2  -  tan theta_1)`

Exercise 6.4 | Q 4 | Page 265

The angle of elevation of the top of a cell phone tower from the foot of a high apartment is 60° and the angle of depression of the foot of the tower from the top of the apartment is 30°. If the height of the apartment is 50 m, find the height of the cell phone tower. According to radiation control norms, the minimum height of a cell phone tower should be 120 m. State if the height of the above mentioned cell phone tower meets the radiation norms

Exercise 6.4 | Q 5. (i) | Page 265

The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 60° and 30° respectively. Find the height of the lamp post

Exercise 6.4 | Q 5. (ii) | Page 265

The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 60° and 30° respectively. Find the difference between height of the lamp post and the apartment

Exercise 6.4 | Q 5. (iii) | Page 265

The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 60° and 30° respectively. Find the distance between the lamp post and the apartment `(sqrt(3) = 1.732)`

Exercise 6.4 | Q 6. (i) | Page 265

Three villagers A, B and C can see each other using telescope across a valley. The horizontal distance between A and B is 8 km and the horizontal distance between B and C is 12 km. The angle of depression of B from A is 20° and the angle of elevation of C from B is 30°. Calculate the vertical height between A and B. (tan 20° = 0.3640, `sqrt3` = 1.732)

Exercise 6.4 | Q 6. (ii) | Page 265

Three villagers A, B and C can see each other using telescope across a valley. The horizontal distance between A and B is 8 km and the horizontal distance between B and C is 12 km. The angle of depression of B from A is 20° and the angle of elevation of C from B is 30°. Calculate the vertical height between B and C. (tan 20° = 0.3640, `sqrt3` = 1.732)

Exercise 6.5 [Pages 265 - 267]

Samacheer Kalvi solutions for Mathematics [English] Class 10 SSLC TN Board 6 Trigonometry Exercise 6.5 [Pages 265 - 267]

Multiple Choice Questions

Exercise 6.5 | Q 1 | Page 265

The value of sin2θ + `1/(1 + tan^2 theta)` is equal to 

  • tan2θ

  • 1

  • cot2θ

  • 0

Exercise 6.5 | Q 2 | Page 265

tan θ cosec2 θ – tan θ is equal to

  • sec θ

  • cot2 θ

  • sin θ

  • cot θ

Exercise 6.5 | Q 3 | Page 265

If (sin α + cosec α)2 + (cos α + sec α)2 = k + tan2α + cot2α, then the value of k is equal to

  • 9

  • 7

  • 5

  • 3

Exercise 6.5 | Q 4 | Page 266

If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b(a2 – 1) is equal to

  • 2a

  • 3a

  • 0

  • 2ab

Exercise 6.5 | Q 5 | Page 266

If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to 

  • 25

  • `1/25`

  • 5

  • 1

Exercise 6.5 | Q 6 | Page 266

If sin θ = cos θ , then 2 tan2θ + sin2θ – 1 is equal to

  • `(-3)/2`

  • `3/2`

  • `2/3`

  • `(-2)/3`

Exercise 6.5 | Q 7 | Page 266

If x = a tan θ and y = b sec θ then

  • `y^2/"b"^2 - x^2/"a"^2` = 1

  • `x^2/"a"^2 - y^2/"b"^2` = 1

  • `x^2/"a"^2 + y^2/"b"^2` = 1

  • `x^2/"a"^2 - y^2/"b"^2` = 0

Exercise 6.5 | Q 8 | Page 266

(1 + tan θ + sec θ) (1 + cot θ − cosec θ) = ______.

  • 0

  • 1

  • 2

  • -1

  • none of these

Exercise 6.5 | Q 9 | Page 266

a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to

  • a2 – b2

  • b2 – a2

  • a2 + b2

  • b – a

Exercise 6.5 | Q 10 | Page 266

If the ratio of the height of a tower and the length of its shadow is `sqrt(3): 1`, then the angle of elevation of the sun has measure

  • 45°

  • 30°

  • 90°

  • 60°

Exercise 6.5 | Q 11 | Page 266

The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘b’ metres above the first, the depression of the foot of the pole is 60°. The height of the pole (in metres) is equal to

  • `sqrt(3)` b

  • `"b"/3`

  • `"b"/2`

  • `"b"/sqrt(3)`

Exercise 6.5 | Q 12 | Page 266

A tower is 60 m heigh. Its shadow is x metres shorter when the sun’s altitude is 45° than when it has been 30°, then x is equal to

  • 41.92 m

  • 43.92 m

  • 43 m

  • 45.6 m

Exercise 6.5 | Q 13 | Page 266

The angle of depression of the top and bottom of 20 m tall building from the top of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between two buildings (in metres) is 

  • `20, 10sqrt(3)`

  • `30, 5sqrt(3)`

  • 20, 10

  • `30, 10sqrt(3)`

Exercise 6.5 | Q 14 | Page 266

Two persons are standing ‘x’ metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is 

  • `sqrt(2)x`

  • `x/(2sqrt(2))`

  • `x/sqrt(2)`

  • 2x

Exercise 6.5 | Q 15 | Page 267

The angle of elevation of a cloud from a point h metres above a lake is β. The angle of depression of its reflection in the lake is 45°. The height of location of the cloud from the lake is

  • `("h"(1 + tan beta))/(1 - tan beta)`

  • `("h"(1 - tan beta))/(1 + tan beta)`

  • h tan(45° − β)

  • none of these

Unit Exercise – 6 [Page 267]

Samacheer Kalvi solutions for Mathematics [English] Class 10 SSLC TN Board 6 Trigonometry Unit Exercise – 6 [Page 267]

Unit Exercise – 6 | Q 1. (i) | Page 267

Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0

Unit Exercise – 6 | Q 1. (ii) | Page 267

Prove that `(tan^2 theta - 1)/(tan^2 theta + 1)` = 1 – 2 cos2θ

Unit Exercise – 6 | Q 2 | Page 267

Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`

Unit Exercise – 6 | Q 3 | Page 267

If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1

Unit Exercise – 6 | Q 4 | Page 267

If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `±  sqrt("a"^2 + "b"^2 -"c"^2)`

Unit Exercise – 6 | Q 5 | Page 267

A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such away 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 30°. Determine the speed at which the bird flies `(sqrt(3) = 1.732)`

Unit Exercise – 6 | Q 6 | Page 267

An aeroplane is flying parallel to the Earth’s surface at a speed of 175 m/sec and at a height of 600 m. The angle of elevation of the aeroplane from a point on the Earth’s surface is 37°. After what period of time does the angle of elevation increase to 53°? (tan 53° = 1.3270, tan 37° = 0.7536)

Unit Exercise – 6 | Q 7. (i) | Page 267

A bird is flying from A towards B at an angle of 35°, a point 30 km away from A. At B it changes its course of flight and heads towards C on a bearing of 48° and distance 32 km away. How far is B to the North of A?

(sin 55° = 0.8192, cos 55° = 0.5736, sin 42° = 0.6691, cos 42° = 0.7431)

Unit Exercise – 6 | Q 7. (ii) | Page 267

A bird is flying from A towards B at an angle of 35°, a point 30 km away from A. At B it changes its course of flight and heads towards C on a bearing of 48° and distance 32 km away. How far is B to the West of A?

(sin 55° = 0.8192, cos 55° = 0.5736, sin 42° = 0.6691, cos 42° = 0.7431)

Unit Exercise – 6 | Q 7. (iii) | Page 267

A bird is flying from A towards B at an angle of 35°, a point 30 km away from A. At B it changes its course of flight and heads towards C on a bearing of 48° and distance 32 km away. How far is C to the North of B?

(sin 55° = 0.8192, cos 55° = 0.5736, sin 42° = 0.6691, cos 42° = 0.7431)

Unit Exercise – 6 | Q 7. (iv) | Page 267

A bird is flying from A towards B at an angle of 35°, a point 30 km away from A. At B it changes its course of flight and heads towards C on a bearing of 48° and distance 32 km away. How far is C to the East of B?
(sin 55° = 0.8192, cos 55° = 0.5736, sin 42° = 0.6691, cos 42° = 0.7431)

Unit Exercise – 6 | Q 8 | Page 267

Two ships are sailing in the sea on either side of the lighthouse. The angles of depression of two ships as observed from the top of the lighthouse are 60° and 45° respectively. If the distance between the ships is `200[(sqrt(3) + 1)/sqrt(3)]` metres, find the height of the lighthouse.

Unit Exercise – 6 | Q 9 | Page 267

A building and a statue are in opposite side of a street from each other 35 m apart. From a point on the roof of building the angle of elevation of the top of statue is 24° and the angle of depression of base of the statue is 34°. Find the height of the statue. (tan 24° = 0.4452, tan 34° = 0.6745)

Solutions for 6: Trigonometry

Exercise 6.1Exercise 6.2Exercise 6.3Exercise 6.4Exercise 6.5Unit Exercise – 6
Samacheer Kalvi solutions for Mathematics [English] Class 10 SSLC TN Board chapter 6 - Trigonometry - Shaalaa.com

Samacheer Kalvi solutions for Mathematics [English] Class 10 SSLC TN Board chapter 6 - Trigonometry

Shaalaa.com has the Tamil Nadu Board of Secondary Education Mathematics Mathematics [English] Class 10 SSLC TN Board Tamil Nadu Board of Secondary Education solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Samacheer Kalvi solutions for Mathematics Mathematics [English] Class 10 SSLC TN Board Tamil Nadu Board of Secondary Education 6 (Trigonometry) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. Samacheer Kalvi textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.

Concepts covered in Mathematics [English] Class 10 SSLC TN Board chapter 6 Trigonometry are Trigonometry, Trigonometric Identities, Heights and Distances.

Using Samacheer Kalvi Mathematics [English] Class 10 SSLC TN Board solutions Trigonometry exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in Samacheer Kalvi Solutions are essential questions that can be asked in the final exam. Maximum Tamil Nadu Board of Secondary Education Mathematics [English] Class 10 SSLC TN Board students prefer Samacheer Kalvi Textbook Solutions to score more in exams.

Get the free view of Chapter 6, Trigonometry Mathematics [English] Class 10 SSLC TN Board additional questions for Mathematics Mathematics [English] Class 10 SSLC TN Board Tamil Nadu Board of Secondary Education, and you can use Shaalaa.com to keep it handy for your exam preparation.

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