CBSE Class 10 Mathematics Some Applications of Trigonometry Notes Set 05

Basic Terms

The line of sight is the straight line from the observer's eye to the object. The angle of elevation is the angle the line of sight makes above the horizontal when the object is higher than the observer. The angle of depression is the angle the line of sight makes below the horizontal when the object is lower than the observer.
Both angles are measured from the horizontal line through the eye, never from the vertical. For a point on the ground seen from the top of a tower, the angle of depression equals the angle of elevation of the top of the tower seen from that point (alternate angles).

Standard Values to Remember

\( \tan 30^\circ = \dfrac{1}{\sqrt{3}} \), \( \tan 45^\circ = 1 \), \( \tan 60^\circ = \sqrt{3} \)
\( \sin 30^\circ = \dfrac{1}{2} \), \( \sin 45^\circ = \dfrac{1}{\sqrt{2}} \), \( \sin 60^\circ = \dfrac{\sqrt{3}}{2} \)
Use \( \sqrt{3} \approx 1.732 \) and \( \sqrt{2} \approx 1.414 \) when a decimal answer is wanted.

Method for Height and Distance Problems

1. Draw a neat figure and mark the horizontal ground line, the vertical height, and the line of sight.
2. Mark the given angles at the observer's position and mark known lengths.
3. Identify the right-angled triangle(s) and choose the ratio that links the known and unknown sides; tan is used most often because it connects height with horizontal distance.
4. Form one equation for a single observation or two equations when the observer moves.
5. Solve, and check that the answer is reasonable (a steeper angle means the observer is nearer the object).

Two Observations from the Same Line

If an observer moves a distance d towards a tower and the angle of elevation changes from \( \alpha \) to \( \beta \) (with \( \beta > \alpha \)), let the height be h. Then the two distances from the foot are \( h\cot\alpha \) and \( h\cot\beta \), so
\( h\cot\alpha - h\cot\beta = d \Rightarrow h = \dfrac{d}{\cot\alpha - \cot\beta} \)
The same idea works for shadows of different lengths at different positions of the sun, and for two ships seen from a lighthouse (add the distances if the ships are on opposite sides, subtract if on the same side).

Worked Example 1: Moving Nearer

From a point on level ground the angle of elevation of a tower is 30°. On moving 20 m towards the tower the angle becomes 60°. Find the height.
Let the height be h. Then \( h\cot 30^\circ - h\cot 60^\circ = 20 \), i.e. \( h\left(\sqrt{3} - \dfrac{1}{\sqrt{3}}\right) = 20 \), so \( h\cdot\dfrac{2}{\sqrt{3}} = 20 \).
Therefore \( h = 10\sqrt{3} \approx 17.32 \) m.

Worked Example 2: Aeroplane Observed from Two Points

The angles of elevation of an aeroplane flying vertically above the ground are 45° and 60° from two points 1 km apart on the same side. Find its height.
Let the height be h km. The distances from the foot are h (at 45°) and \( h/\sqrt{3} \) (at 60°), so \( h - \dfrac{h}{\sqrt{3}} = 1 \).
Hence \( h = \dfrac{\sqrt{3}}{\sqrt{3} - 1} = \dfrac{3 + \sqrt{3}}{2} \approx 2.37 \) km.

Worked Example 3: Tower with a Flagstaff

From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed on a 20 m building are 45° and 60°. Find the height of the tower.
The distance from the point to the building is \( 20\cot 45^\circ = 20 \) m. The top of the tower is at height \( 20\tan 60^\circ = 20\sqrt{3} \) m above the ground.
So the tower is \( 20\sqrt{3} - 20 = 20(\sqrt{3} - 1) \approx 14.64 \) m high.

Common Mistakes

1. Measuring the angle from the vertical instead of from the horizontal.
2. Forgetting the height of the observer (eye level) when the observer is tall, for example a person 1.5 m tall looking at a chimney.
3. Adding distances when they should be subtracted (both observations on the same side of the object).
4. Using the wrong ratio: for a height h and horizontal distance x use tan, not sin or cos, unless the slant length is involved.
5. Rounding \( \sqrt{3} \) too early; keep the surd form until the last step.