RD Sharma Class 10 Solutions Chapter 12- Some Applications of Trigonometry Exercise 12.1 (Updated for 2024)

RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1: Students can download the RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1 PDF to learn how to solve the questions in this exercise correctly. Students wishing to brush up on their concepts can check the RD Sharma Class 10 Solutions.

Download RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1 Free PDF

RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1

Access Answers to Maths RD Sharma Solutions for Class 10 Chapter 12 – Some Applications Of Trigonometry

Exercise 12.1 Page No: 12.29

1. A tower stands vertically on the ground. From a point on the ground, 20 m away from the foot of the tower, the angle of elevation of the top of the tower is 60°. What is the height of the tower?

Solution:

Given:

Distance between the foot of the tower and point of observation = 20 m = BC

Angle of elevation of the top of the tower = 60° = θ

And, Height of tower (H) = AB

Now, from fig. ABC

ΔABC is a right-angle triangle,

So,

2. The angle of elevation of a ladder against a wall is 60° and the foot of the ladder is 9.5 m away from the wall. Find the length of the ladder.

Solution:

Given:

Distance between the wall and foot of the ladder = 9.5 m

Angle of elevation (θ) = 60°

Length of the ladder = L = AC

Now, from fig. ABC

ΔABC is a right angle triangle,

So,

Thus, length of the ladder (L) = 19 m

3. A ladder is placed along a wall of a house such that its upper end is touching the top of the wall. The foot of the ladder is 2 m away from the wall and the ladder is making an angle of 60° with the level of the ground. Determine the height of the wall.

Solution:

Given,

Distance between the wall and the foot of the ladder = 2m = BC

Angle made by ladder with ground (θ) = 60°

Height of the wall (H) = AB

Now, the fig. of ABC forms a right angle triangle.

So,

4. 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.

Solution:

Given,

Height of the electric pole = 10 m = AB

The angle made by steel wire with the ground (horizontal) θ = 45°

Let length of wire = L = AC

So, from the figure formed we have ABC as a right triangle.

5. A kite is flying at a height of 75 meters from the ground level, attached to a string inclined at 60° to the horizontal. Find the length of the string to the nearest meter.

Solution:

Given,

Height of kite flying from the ground level = 75 m = AB

Angle of inclination of the string with the ground (θ) = 60°

Let the length of the string be L = AC

So, from the figure formed we have ABC as a right triangle.

Hence,

6. A ladder 15 metres long reaches the top of a vertical wall. If the ladder makes an angle of 60^o with the wall, find the height of the wall.

Solution:

Given,

The length of the ladder = 15m = AO

Angle made by the ladder with the wall = 60^o

Let the height of the wall be h metres.

And the horizontal ground taken as OX.

Then from the fig. we have,

In right ΔABO, using trigonometric ratios

cos (60^o) = AB/AO

1/2 = h/ 15

h = 15/2

h = 7.5m

Hence, the height of the wall is 7.5m

7. A vertical tower stands on a horizontal place and is surmounted by a vertical flag staff. At a point on the plane 70 meters away from the tower, an observer notices that the angles of elevation of the top and bottom of the flag-staff are respectively 60° and 45°. Find the height of the flag staff and that of the tower.

Solution:

Given,

A vertical tower is surmounted by flag staff.

Distance between observer and the tower = 70 m = DC

Angle of elevation of bottom of the flag staff = 45°

Angle of elevation of top of the flag staff = 60°

Let the height of the flag staff = h = AD

Height of tower = H = BC

If we represent the above data in the figure then it forms right angle triangles ΔACD and ΔBCD

When θ is angle in right angle triangle we know that

tan θ = opp. Side/ Adj. side

Now,

tan 45^o = BC/ DC

1 = H/ 70

∴ H =70 m

Again,

x = 70 (1.732-1)

∴ x = 51.24 m

Therefore, the height of tower = 70 m and the height of flag staff = 51.24 m

8. A vertically straight tree, 15 m high, is broken by the wind in such a way that its top just touches the ground and makes an angle of 60° with the ground. At what height from the ground did the tree break?

Solution:

Given,

The initial height of tree H = 15 m = AB + AC

Let us assume that it is broken at point A.

And, the angle made by broken part with the ground (θ) = 60°

Height from ground to broken points = h = AB

So, we have

H = AC + h

⟹ AC = (H – h) m

We get a right triangle formed by the above given data,

So,

Rationalizing denominator, we have

Therefore, the height of broken point from the ground is 15(2√3 – 3)m

9. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height 5 meters. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are respectively 30° and 60°. Find the height of the tower.

Solution:

Given,

Height of the flag staff = 5 m =AB

Angle of elevation of the top of flag staff = 60°

Angle of elevation of the bottom of the flagstaff = 30°

Let height of tower be ‘h’ m = BC

And, let the distance of the point from the base of the tower = x m

In right angle triangle BCD, we have

tan 30^o = BC/DC

1/√3 = h/x

x = h√3 ….. (i)

Now, in ΔACD,

tan 60^o = AC/DC

√3 = (5 + h)/ x

√3x = 5 + h

√3(h√3) = 5 + h [using (i)]

3h = 5 + h

2h = 5

h = 5/2 = 2.5m

Therefore, the height of the tower = 2.5 m

10. A person observed the angle of elevation of a tower as 30°. He walked 50 m toward the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60°. Find the height of the tower.

Solution:

Given,

The angle of elevation of the tower before he started walking = 30^o

Distance walked by the person towards the tower = 50m

The angle of elevation of the tower after he walked = 60^o

Let height of the tower (AB) = h m

Let the distance BC = x m

From the fig. in ΔABC,

tan 60^o = AB/ BC

√3 = h/x

x = h/√3 ….(i)

Now, in ΔABD

tan 30^o = AB/ BD

1/√3 = h/ (50 + x)

√3h = 50 + x

√3h = 50 + (h/√3) [using (i)]

3h = 50√3 + h

2h = 50√3

h = 25√3 = 25(1.73) = 43.25m

Therefore, the height of the tower = 43.25m

11. The shadow of a tower, when the angle of elevation of the sun is 45°, is found to be 10 m longer than when it was 60°. Find the height of the tower.

Solution:

Let the height of the tower(AB) = h m

Let the length of the shorter shadow be x m

Then, the longer shadow is (10 + x)m

So, from fig. In ΔABC

tan 60^o = AB/BC

√3 = h/x

x = h/√3…. (i)

Next, in ΔABD

tan 45^o = AB/BD

1 = h/(10 + x)

10 + x = h

10 + (h/√3) = h [using (i)]

10√3 + h = √3h

h(√3 -1) =10√3

h = 10√3/ (√3 -1)

After rationalising the denominator, we have

h = [10√3 x (√3 + 1)]/ (3 – 1)

h = 5√3(√3 + 1)

h = 5(3 + √3) = 23.66 [√3 = 1.732]

Therefore, the height of the tower is 23.66 m.

12. A parachute is descending vertically and makes angles of elevation of 45° and 60° at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of point where he falls on the ground from the just observation point.

Solution:

Let the parachute at highest point A and let C and D be points which are 100 m apart on ground where from then CD = 100 m

Angle of elevation from point D = 45° = α

Angle of elevation from point C = 60° = β

Let B be the point just vertically down the parachute

Let us draw figure according to above data then it forms the figure as shown in which ABC and ABD are two triangles

Maximum height of the parachute from the ground

AB = H m

Distance of point where parachute falls to just nearest observation point = x m

If in right angle triangle one of the included angles is θ then

From (a) and (b)

x = 136.6 m in (b)

H = √3 × 136.6 = 236.6m

Therefore,

The maximum height of the parachute from the ground, H = 236.6m

The distance between the two points where the parachute falls on the ground and just the observation is x = 136.6 m

We have provided complete details of RD Sharma Class 10 Maths Solutions Chapter 12 Exercise 12.1. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.

FAQs on RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1