# Notes of Ch 9 Some Applications of Trigonometry| Class 10th Math

## Revision Notes for Ch 9 Some Application of Trigonometry Class 10th Mathematics

**Introduction**

→ This chapter is an application of trigonometry. In trigonometry, we use to measure sides of triangle, when particular angle was given. Most the buildings, walls, towers we see around are perpendicular to ground. The chapter deals with measurement of heights and distance from certain points with the help of trigonometry if particular angles are known.

**Line of Sight:**When an observer looks from a point O at an object A, then OA is called line of sight.

**Angle of Elevation:**Suppose that from a point O, you look up at point A, then the angle, the which the line of sight makes with the horizontal line through O is called the angle of elevation of A, as we seen from O.

Here OX is horizontal line. is here angle of elevation.

**Angle of depression:**Suppose that from the point O, you look down at an object B, placed below the level of your eye. Then the angle which the line of sight makes with the horizontal line through O is called angle of depression of B for O.

Here, OX is horizontal line. β is here angle of deviation.

**An Example of Angle of Elevation and Depression**

A man standing on a building will look at a building higher than his. If he sees at the top, he will look at the top he will form angle of elevation. If he look at the bottom he will form angle of depression with bottom of the building.

**Solved Example based on The Concept**

Q. The angle of elevation of a cloud 60m above a lake is 30° and the angle of reflection of the cloud in the lake is 60°. Find height of the cloud.

**Answer**

The situation is depicted by the following diagram.

Let AB be the surface of the lake and let P be the point vertically above A such that AP=60m

Let C be the position of cloud and let D be its reflection in the lake.

Let the height of Cloud be H meters.

Then BC=BD=H (Let)

Draw PQ Perpendicular to CD

⇒∠QPC=30°,∠QPD=60°

BQ=AP=60m

CQ=(H-60)m,DQ=(H+60)m

From right triangle ∆ CQP, we have

Let AB be the surface of the lake and let P be the point vertically above A such that AP=60m

Let C be the position of cloud and let D be its reflection in the lake.

Let the height of Cloud be H meters.

Then BC=BD=H (Let)

Draw PQ Perpendicular to CD

⇒∠QPC=30°,∠QPD=60°

BQ=AP=60m

CQ=(H-60)m,DQ=(H+60)m

From right triangle ∆ CQP, we have

PQ/CQ=cot30°

PQ/(H-60) = √3

PQ⇒(H-60) √3 m (i)

From the right triangle ∆ DQP

PQ/DQ=cot60°

⇒PQ=(H+60)/√3 (ii)

From (i) and (ii), we get

(H-60) √3=(H+60)/√3

⇒3H-180=H+60

⇒H=240

So, height of the cloud is H=240m.

**Ladder Problem**

**Answer**

The angle of elevation of a ladder leaning against a wall is 45° and foot of the ladder is 10m away from the wall. Find the length of ladder. The following diagram depicts the situation

Let OY is the ladder. Let OY=l. YX is the wall on which ladder is leaning on. It is given, ladder is leaned at 45°. Also, distance from the wall to foot of ladder is 10m, OX=d=10m. It is given √2=1.414

From the diagram,

cos 45°= OX/OY = d/l

⇒ l=d/cos 45°

We know that cos 45° = 1/√2

So, l = d/(1/√2) = √2 d

l = √2×10 m = 10√2 m = 10×1.414 = 14.14m

So, length of ladder is 14.14m