NIMCET Heights And Distances Previous Year Questions (PYQs) – Page 2 of 2

By sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

So,

$\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A+\sin B}$

Now,

$\sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

and

$\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

Therefore,

$\frac{\sin A-\sin B}{\sin A+\sin B}=\cot\left(\frac{A+B}{2}\right)\tan\left(\frac{A-B}{2}\right)$

Since,

$A+B=\pi-C$

So,

$\cot\left(\frac{A+B}{2}\right)=\cot\left(\frac{\pi-C}{2}\right)$

$=\cot\left(\frac{\pi}{2}-\frac{C}{2}\right)$

$=\tan\left(\frac{C}{2}\right)$

Thus,

$\frac{a-b}{a+b}=\tan\left(\frac{C}{2}\right)\tan\left(\frac{A-B}{2}\right)$

But option $4$ gives

$\frac{a-b}{a+b}=\frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{C}{2}\right)}$

Let distances from the lighthouse be $x$ (for $30^\circ$) and $y$ (for $45^\circ$). With height $h=100$ m:

$\tan 30^\circ=\dfrac{100}{x}\ \Rightarrow\ x=\dfrac{100}{\tan30^\circ}=100\sqrt{3}$,   $\tan 45^\circ=\dfrac{100}{y}\ \Rightarrow\ y=\dfrac{100}{\tan45^\circ}=100$.

Ships are on opposite sides ⇒ distance $=x+y=100\sqrt{3}+100=100(\sqrt{3}+1)\approx \boxed{273.2\ \text{m}}$.