Some Applications of Trigonometry — RBSE Class 10 (Mathematics)

How tall is that tower? How wide is the river? You cannot climb it or swim it — but stand back, measure an angle, and trigonometry hands you the answer. This chapter is a single, very practical idea: turn a real scene into a right triangle, then let a ratio do the measuring.

1. Line of sight, elevation and depression

• Line of sight — the straight line from your eye to the object.
• Angle of elevation — the angle above the horizontal when you look up at something higher.
• Angle of depression — the angle below the horizontal when you look down at something lower.

Key fact used constantly: the angle of depression from the top of a tower to a point on the ground equals the angle of elevation from that point to the top (they are alternate angles between parallel horizontals).

2. The three ratios, in this setting

Set up the right triangle so the unknown is a side; then pick the ratio linking the unknown, the known side, and the angle:

$\tan \theta =\frac{\text{opposite}}{\text{adjacent}},\sin \theta =\frac{\text{opposite}}{\text{hyp}},\cos \theta =\frac{\text{adjacent}}{\text{hyp}}$

Most height-and-distance problems use tan (height vs horizontal distance). Use the standard angles 30°, 45°, 60°:

3. A reliable method

1. Draw the scene as a right triangle; mark the horizontal, the vertical, and the angle.
2. Label what is known and what is asked.
3. Choose the ratio that involves exactly those two sides and the angle.
4. Solve, and rationalise surds if needed (e.g. $\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$).

Example — the angle of elevation of the top of a tower from a point 30 m away is 60°. Height? $\tan 60°=\frac{h}{30}\Rightarrow h=30\sqrt{3}\approx 51.96\text{ m}.$

4. Two-triangle problems

The longer questions use two right triangles sharing a side — e.g. angles of elevation from two points, or elevation to the top and depression to the base. Form one equation per triangle and eliminate the shared unknown.

Example idea — from a point the elevation of a tower's top is 30°; walking 20 m nearer it becomes 60°. Two tan-equations in the height $h$ and the far distance give $h=10\sqrt{3}$ m.

5. Practical cautions

• If the observer has a height (a boy 1.5 m tall), the triangle's vertical side is (object height − observer height); add the observer's height back at the end.
• Keep answers exact with surds unless a decimal is asked (take $\sqrt{3}\approx 1.732$).

6. Closing thought

There is really one skill here: convert words into a right triangle and choose tan, sin or cos. With the standard-angle values memorised, these become some of the fastest marks in the paper. The RBSE board almost always sets one height-and-distance problem — frequently the two-triangle type — so practise drawing the figure quickly and cleanly.