CBSE Class 10 Mathematics Some Applications of Trigonometry Notes Set 06

Line of Sight, Elevation and Depression

The line of sight is the line from the observer's eye to the object being viewed.
The angle of elevation is the angle between the horizontal through the observer's eye and the line of sight when the object is above the horizontal.
The angle of depression is the angle between the horizontal through the observer's eye and the line of sight when the object is below the horizontal.
The angle of depression of an object from a height equals the angle of elevation of the observer from the object (alternate angles between parallel horizontals).

Method of Solving

1. Draw a clear figure, marking the heights, distances and given angles.
2. Identify the right triangles and decide which side is opposite and which is adjacent to the given angle.
3. Choose the ratio that links the known and unknown sides: tan θ = height/base is used most often; use sin θ or cos θ when the hypotenuse (ladder, rope, wire) is involved.
4. Solve, and keep exact values such as √3 until the final step, using √3 = 1.732 and √2 = 1.414 if numerical values are asked.

Standard Results

tan 30° = 1/√3, tan 45° = 1, tan 60° = √3.
1. If the angle of elevation changes from x to y when the observer moves a distance d nearer the tower, then the height h = d/(cot x − cot y) = d tan x tan y/(tan y − tan x).
2. If the angles of elevation from two points at distances a and b from the foot (on the same side) are complementary, then h = √(ab).
3. A tree broken at height h whose top touches the ground at distance d from the root: length of the broken part = √(h² + d²) and the original height is h + √(h² + d²).
4. A ladder of length l leaning against a wall at angle θ with the ground reaches height l sin θ, and its foot is at distance l cos θ.
5. For two poles of heights p and q seen from the midpoint of the line joining their feet at elevations α and β, the horizontal distance is the same, so p/q = tan α/tan β.

Worked Examples

Example 1. A tower stands vertically on the ground. From a point 15 m from its foot, the angle of elevation of its top is 60°. Find the height of the tower.
Solution: h = 15 tan 60° = 15√3 ≈ 25.98 m.

Example 2. A kite is flying at a height of 60 m above the ground. The string attached to it makes an angle of 60° with the level ground. Find the length of the string, assuming it is straight.
Solution: sin 60° = 60/l, so l = 60/(√3/2) = 40√3 ≈ 69.28 m.

Example 3. From the top of a 20 m high building the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.
Solution: Horizontal distance = 20 cot 45° = 20 m. Part above the building = 20 tan 60° = 20√3. Height = 20 + 20√3 = 20(√3 + 1) ≈ 54.64 m.

Example 4. The angle of elevation of the top of a tower from two points 4 m and 9 m from its base, in the same line, are complementary. Find the height.
Solution: h = √(4 × 9) = 6 m.

Example 5. From the top of a hill, the angles of depression of two consecutive kilometre stones, due east, are 30° and 45°. Find the height of the hill.
Solution: h cot 30° − h cot 45° = 1, so h(√3 − 1) = 1 and h = (√3 + 1)/2 ≈ 1.366 km.