CBSE Class X

Question 11:

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

Answer:

In ΔABC,

Therefore, the height of the tower is 10√3 m and the width of the canal is 10 m.

Question 10:

Two poles of equal heights are standing opposite each other an either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30º, respectively. Find the height of poles and the distance of the point from the poles.

Answer:

Let AB and CD be the poles and O is the point from where the elevation angles are measured.

In ΔABO,

Since the poles are of equal heights,

CD = AB

DO = BD − BO = (80 − 20) m = 60 m

Therefore, the height of poles is 20√3 and the point is 20 m and 60 m far from these poles.

Question 9:

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer:

Let AB be the building and CD be the tower.

In ΔCDB,

Therefore, the height of the building is .

Question 8:

A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45 °. Find the height of the pedestal.

Answer:

Let AB be the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured.

Therefore, the height of the pedestal is 0.8 (√3 + 1)m.

Question 7:

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer:

Let BC be the building, AB be the transmission tower, and D be the point on the ground from where the elevation angles are to be measured.

In ΔBCD,

Therefore, the height of the transmission tower is 20(√3 – 1) m.

Question 6:

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer:

Let the boy was standing at point S initially. He walked towards the building and reached at point T.

It can be observed that

PR = PQ − RQ

= (30 − 1.5) m = 28.5 m = 57/2 m

Hence, he walked 19√3 m towards the building.

Question 5:

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer:

Let K be the kite and the string is tied to point P on the ground.

Hence, the length of the string is 40√3 m.

Question 4:

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 the tower is 30°. Find the height of the tower.

Answer:

Let AB be the tower and the angle of elevation from point C (on ground) is 30°.

In ΔABC,

Therefore, the height of the tower is 10 √3 m.

Question 3:

A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30 ° to the ground, where as for the elder children she wants to have a steep side at a height of 3 m, and inclined at an angle of 60 ° to the ground. What should be the length of the slide in each case?

Answer:

It can be observed that AC and PR are the slides for younger and elder children respectively.

Therefore, the lengths of these slides are 3 m and 2√3 m.

Question 2:

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30 ° 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.

Answer:

Let AC was the original tree. Due to storm, it was broken into two parts. The broken part A’B making 30° with the ground.

Height of tree = A’B + BC

Hence, the height of the tree is 8√3 m.

Question 1:

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30 °.

Answer:

It can be observed from the figure that AB is the pole.

In ‘ABC,

Therefore, the height of the pole is 10 m.

Question 2:

Write all the other trigonometric ratios of ∠A in terms of sec A.

Answer:

We know that,

cos A= 1/sec A

Also, sin² A + cos² A = 1

sin² A = 1 − cos² A

Question 5:

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer:

= secθ cosec θ + 1

= R.H.S.

 

 

 

Question 4:

Choose the correct option. Justify your choice.

(i) 9 sec² A − 9 tan² A =

(A) 1               (B) 9                  (C) 8                 (D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ)

(A) 0            (B) 1                     (C) 2                 (D) −1

(iii) (secA + tanA) (1 − sinA) =

(A) secA     (B) sinA              (C) cosecA       (D) cosA

(A) sec² A    (B) −1                (C) cot² A        (D) tan² A

Answer:

(i) 9 sec² A − 9 tan² A

= 9 (sec²A − tan²A)

= 9 (1) [As sec² A − tan² A = 1]

= 9

Hence, alternative (B) is correct.

(ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ)

Hence, alternative (C) is correct.

(iii) (secA + tanA) (1 − sinA)

Hence, alternative (D) is correct.

Hence, alternative (D) is correct.

Question 3:

Evaluate

(ii) sin25° cos65° + cos25° sin65°

Answer:

(ii) sin25° cos65° + cos25° sin65°

= (sin25°) {cos (90°-25°)} + cos 25° {sin (90° – 25°)}

= (sin 25°) (sin 25°) + (cos 25°) (cos 25°)

= = sin²25° + cos²25°

= = 1 (As sin²A + cos²A = 1)

Question 1:

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Answer:

We know that,

Question 7:

Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Answer:

sin 67° + cos 75°

= sin (90° − 23°) + cos (90° − 15°)

= cos 23° + sin 15°

Question 6:

If A, Band C are interior angles of a triangle ABC then show that

Answer:

We know that for a triangle ABC,

∠ A + ∠B + ∠C = 180°

             ∠B + ∠C= 180° − ∠A

 

Question 5:

If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.

Answer:

Given that,

sec 4A = cosec (A − 20°)

cosec (90° − 4A) = cosec (A − 20°)

90° − 4A= A− 20°

110° = 5A

A = 22°

Question 4:

If tan A = cot B, prove that A + B = 90°

Answer:

Given that,

tan A = cot B

tan A = tan (90° − B)

A = 90° − B

A + B = 90°