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

RD Sharma Class 10 Solutions Chapter 12 Some Applications of Trigonometry: Students can use RD Sharma Solutions for Class 10 Maths Chapter 12 to study and prepare for their board exams. It just has one set of problems focusing on finding heights and distances using trigonometric results. The RD Sharma Solutions for Class 10 is the place to go if you want to learn the correct step-by-step technique and approach to these questions.

Download RD Sharma Class 10 Solutions Chapter 12 Free PDF

RD Sharma Class 10 Solutions Chapter 12 PDF- Direct Link

Exercise-Wise RD Sharma Class 10 Maths Solutions Chapter 12 – Some Applications of Trigonometry

RD Sharma Class 10 Solutions Chapter 12 Exercises

RD Sharma Solutions for Class 10 Chapter 12 Exercise 12.1

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

 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 the point of observation = 20 m = BC

The 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

The angle of elevation (θ) = 60°

Length of the ladder = L = AC

Now, from fig. ABC

ΔABC is a right-angle triangle,

So,

Thus, the 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 the top of the pole is affixed at a point on the ground to keep the pole upright. 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 the 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

The 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 meters 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 meters.

And the horizontal ground is 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 plane and is surmounted by a vertical flagstaff. 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 flagstaff and that of the tower.

Solution:

Given,

A vertical tower is surmounted by flagstaff.

Distance between observer and the tower = 70 m = DC

The angle of elevation of the bottom of the flagstaff = 45°

The angle of elevation of the top of the flagstaff = 60°

Let the height of the flagstaff = 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 the 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 the tower = 70 m and the height of flagstaff = 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 flagstaff = 5 m =AB

The angle of elevation of the top of flagstaff = 60°

The angle of elevation of the bottom of the flagstaff = 30°

Let the height of the 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 towards 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 the 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

Detailed Exercise Explanation for RD Sharma Class 10 Maths Solutions Chapter 12

We jotted all the exercise-wise solutions down below, for you to have a quick idea of the notions that you will study in detail in each exercise. Let’s get to it-

RD Sharma Class 10 Solutions Chapter 12 Exercise 12.1

This introductory section begins with the basics that students learned in the previous chapter as well. Take for instance the last chapter covered trigonometric ratios. This chapter will brief you on various simple applications of Trigonometry in everyday life. Being one of the ancient mathematical subjects the topics are studied all around the globe.

Moreover, trigonometry finds its application in various disciplines too other than maths. Hence students must consider taking a guidebook such as RD Sharma Class 10 Maths Solutions Chapter 12 PDF for securing a robust knowledge on the subject, which will help them grow for future higher studies too.

Download RD Sharma CBSE Class 10 Maths Chapter 12 Solutions to boost your preparations for the exam. If you have any queries, feel free to ask us in the comment section.

Frequently Asked Questions on RD Sharma Class 10 Solutions Chapter 12