# Help in trignometry! (?)

## Calculator is not allowed in exams. I could find a sin inverse relation but that can't be solved without calculator.

Jan 20, 2018

75°

#### Explanation:

The above figure explains the situation described in the given problem.

$Q \to \text{the starting point}$

$P \to \text{the point at a distance d m from from Q}$

$R \to \text{the point from which angle of depression of P is } {30}^{\circ}$

$S \to \text{the point from which angle of depression of P is } {60}^{\circ}$

Let the angle of inclination of slope of the mountain be $\theta$

Here $Q R = c \mathmr{and} Q S = 2 c$

So RU=csintheta ;QU=c costheta

and ST=2csintheta ;QT=2c costheta

Now $\frac{R U}{P U} = \tan {30}^{\circ}$

$\implies \frac{R U}{Q U + Q P} = \tan {30}^{\circ}$

$\implies \frac{c \sin \theta}{c \cos \theta + d} = \frac{1}{\sqrt{3}}$

$\implies c \cos \theta + d = \sqrt{3} c \sin \theta \ldots \ldots \left[1\right]$

Again $\frac{S T}{P T} = \tan {45}^{\circ}$

$\implies \frac{S T}{Q T + Q P} = \tan {45}^{\circ}$

$\implies \frac{2 c \sin \theta}{2 c \cos \theta + d} = 1$

$\implies 2 c \cos \theta + d = 2 c \sin \theta \ldots \ldots \left[2\right]$

Subtracting (1) from (2) we get

$2 c \sin \theta - \sqrt{3} c \sin \theta = c \cos \theta$

$\implies c \sin \theta \left(2 - \sqrt{3}\right) = c \cos \theta$

$\implies \tan \theta = \frac{1}{2 - \sqrt{3}} = 2 + \sqrt{3}$

$\implies \theta = {75}^{\circ}$

Alternative method

In $\Delta P Q R , \angle Q R P = \left(\theta - {30}^{\circ}\right)$

So $\frac{c}{\sin {30}^{\circ}} = \frac{d}{\sin} \left(\theta - {30}^{\circ}\right)$

$\implies 2 c = \frac{d}{\sin} \left(\theta - {30}^{\circ}\right)$

And in $\Delta P Q S , \angle Q S P = \left(\theta - {45}^{\circ}\right)$

So $\frac{2 c}{\sin {45}^{\circ}} = \frac{d}{\sin} \left(\theta - {30}^{\circ}\right)$

$\implies 2 \sqrt{2} c = \frac{d}{\sin} \left(\theta - {45}^{\circ}\right)$

Hence $\sqrt{2} \sin \left(\theta - {45}^{\circ}\right) = \sin \left(\theta - {30}^{\circ}\right)$

$\implies \sin \theta - \cos \theta = \frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta$

$\implies 2 \sin \theta - 2 \cos \theta = \sqrt{3} \sin \theta - \cos \theta$

$\implies \tan \theta = \frac{1}{2 - \sqrt{3}} = 2 + \sqrt{3} = \tan {75}^{\circ}$

Please note

$\tan {75}^{\circ} = \tan \left({45}^{\circ} + {30}^{\circ}\right)$

$= \frac{\tan {45}^{\circ} + \tan {30}^{\circ}}{1 - \tan {45}^{\circ} \tan {30}^{\circ}}$

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

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

$= {\left(\sqrt{3} + 1\right)}^{2} / \left({\left(\sqrt{3}\right)}^{2} - {1}^{2}\right)$

$= \frac{4 + 2 \sqrt{3}}{2} = 2 + \sqrt{3}$