Question

Two poles are 10 m and 20 m high. The line joining their tips makes an angle of ${15}^o$ with the horizontal. What is the distance between the poles ?

• A. $10(\sqrt{3}-1)m$
• B. $5(4+2\sqrt{3}-1)m$
• C. $20(\sqrt{3}+1)m$
• D. $10(\sqrt{3}+2)m$

Answer 1

The correct option is D $10(\sqrt{3}+2)m$

Now AC will be the distance between the poles

Here,

$\tan \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1-cos\theta }{1+cos\theta }}$

Now, $\tan \left({15}^o\right)=\frac{10}{AC}$

$\Rightarrow AC=\frac{10}{\tan \left({15}^o\right)}$

But, $\tan \left({15}^o\right)=\sqrt{\frac{1-\cos \left({30}^o\right)}{1+\cos \left({30}^o\right)}}$

$=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}}}=\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}}=2-\sqrt{3}$

So, $AC=\frac{10}{2-\sqrt{3}}=\frac{10}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=20+10\sqrt{3}$