Question

A person of height $2m$ wants to get a fruit which is on the top of a pole of height $\frac{10}{3}m$ m if he stands at distance of $\left(\frac{4}{\sqrt{3}}\right)m$ from the foot of the pole, then the angle at which he should throw the stone, so that it hits the fruit is

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${30}^{\circ }$

In the above figure, $ED$ is the man trying to get the fruit on the pole $AC$ and $BC=DE$ distance of the man from the foot of the pole.

Let $\theta$ be the angle at which the man should throw the stone.

From the figure,

$\begin{array}{cc}AC& =AB+BC\\ \frac{10}{3}& =AB+2\\ AB& =\frac{10}{3}-2\\ & =\frac{10-6}{3}\\ & =\frac{4}{3}m\end{array}$

Now, in $△ABE,$

$\begin{array}{cc}\tan \theta & =\frac{AB}{BE}\\ & =\frac{\frac{4}{3}}{\frac{4}{\sqrt{3}}}\\ & =\frac{4}{3}\times \frac{\sqrt{3}}{4}\\ & =\frac{1}{\sqrt{3}}\\ \tan \theta & =\tan {30}^{\circ }\\ \theta & ={30}^{\circ }\end{array}$

Hence, man should throw the stone at ${30}^{\circ }$ to get the fruit from the pole.