First slide

Applications of trigonometry

Question

The angle of elevation of the summit of a mountain from a point on the ground is 45°. After climbing up one km towards the summit at an inclination of 30° from the ground, the angle of elevation of the summit is found to be 60°. Then the height (in km) of the summit from the ground is:

Moderate

A

$\frac{1}{\sqrt{3} - 1}$
B

$\frac{1}{\sqrt{3} + 1}$
C

$\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
D

$\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

Solution

$In Δ BCD$
$\begin{array}{l} \frac{C D}{S in B} = \frac{B C}{Sin D} ∣ \\ \Rightarrow \frac{1}{Sin 15 °} = \frac{B C}{Sin 150 °} & \Rightarrow B C = \frac{\sqrt{2}}{\sqrt{3} - 1} \\ In Δ A B C, Sin 45 ° = \frac{A B}{B C} & \Rightarrow A B = \frac{\sqrt{2}}{\sqrt{3} - 1} \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{3} - 1} \end{array}$

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