Question

The angle of elevation of a tower at a level ground is ${30}^o$ . The angle of elevation becomes $\theta$ when moved 10 m towards the tower. If the height of tower is $5\sqrt{3}m$, then what is $\theta$ equal to ?

• A. ${45}^o$
• B. ${60}^o$
• C. ${75}^o$
• D. None of the above

Answer 1

The correct option is B ${60}^o$

In $△ABC,$

$\tan \left({30}^o\right)=\frac{5\sqrt{3}}{10+x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{5\sqrt{3}}{10+x}$

$\Rightarrow x+10=15$

$\Rightarrow x=5$

In $△ADC,$

$\tan \left(\theta \right)=\frac{5\sqrt{3}}{x}=\frac{5\sqrt{3}}{5}=\sqrt{3}$

$\Rightarrow \theta ={60}^o$

Hence, option B is correct.