Table of Contents

## Heights and Distances in Trigonometry:

Knowledge of Trigonometry can be used to calculate heights of tall objects or distance of inaccessible objects. The heights of a mountain, breadth of a river etc., are generally estimated with the help of trigonometric knowledge. In this connection, a few terms are very important.

The bearing of a Point P is indicated by the magnitude of the angle specifying whether the line segment joins the intersection of the NS and WE lines and the point P is measured from ON or OS and whether it is to the East or West. In **fig. (i)**, the line segment OP is in the direction of 40° to the East of North and is specified as N 40° E.

### Angle Of Elevation:

When the object is at a higher level than that of the observer, then in order to have a view of the object, the observer has to look up. The angle that the line of sight in this case makes with the horizontal through the observer’s eye is known as the angle of elevation. In **fig. (ii)**, θ is the angle of elevation.

### Angle of Depression:

When the object is at a lower level than that of the observer, then the observer has to look below in order to see the object. The angle that the line of sight in this case makes with the horizontal line through the observer’s eye is known as the angle of depression. In** fig. (iii)**, Φ is the angle of depression.

**Line of Sight-** Suppose we are viewing an object standing on the ground. Clearly, the line of sight (or line of vision) to the object is the line from our eyes to the object, we are viewing.

**Example-** A round balloon of radius** **r subtends an angle α at the eye of the observer, while the angle of elevation of its centre is β. Prove that the height of the centre of the balloon is (r sin β cosec α/2).

**Proof:**

Let us represent the balloon by a circle with centre C and radius r. Let OX be the horizontal ground and let O be the point of observation. From O, draw tangents OA nad OB to the circle. Join CA, CB and CO. Draw CD ⊥ OX. Then, ∠AOB = α,

∠AOC = ∠BOC = α/2, ∠ODC = 90° and ∠DOC = β Clearly, ∠OAC = 90°, and CD = height of the centre. From the right △OAC, we have OC/AC = cosec α/2 ⇒ OC/r = cosec α/2 ⇒ OC = r (cosec α/2 ) From the right △ODC, we have CD/OC = sin β ⇒ CD/r(cosec α/2) = sin β ⇒ CD = r sin β (cosec α/2) Hence, the height of the centre of the balloon from the ground is r sin β cosec α/2. |