Matrix Matching : A person walking along a straight road towards a hill observes at two points distance  $\sqrt{3}km$.  The angle of elevation of the hill to be  $30°\text{\hspace{0.17em}and\hspace{0.17em}}60°$.

B) In $\Delta ACB$, $\tan 60°=\frac{AB}{AC}$

$\Rightarrow \sqrt{3}=\frac{h}{x}$

$\Rightarrow \sqrt{3}x=h\to \left(1\right)$

In $\Delta ADB$, $\tan 30°=\frac{AB}{AD}$

$\frac{1}{\sqrt{3}}=\frac{h}{AC+CD}$

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

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+\sqrt{3}}$ (from (1))

$\Rightarrow 3x=x+\sqrt{3}$

$\Rightarrow 2x=\sqrt{3}$

$\Rightarrow x=\frac{\sqrt{3}}{2}$

$\therefore AC=\frac{\sqrt{3}}{2}km$

A) Put  $x=\frac{\sqrt{3}}{2}$ in equation (1)

$\therefore h=\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)$

$h=\frac{3}{2}$

$\therefore$ Height of the hill is $\frac{3}{2}km$.

C) In $\Delta ACB$, $\cos 60°=\frac{AC}{BC}$

$\Rightarrow \frac{1}{2}=\frac{\frac{\sqrt{3}}{2}}{BC}$

$\Rightarrow BC=\frac{\sqrt{3}}{2}\times 2$

$\therefore BC=\sqrt{3}$

$\therefore$ Distance between $B\text{\hspace{0.17em}and\hspace{0.17em}}C$ is  $\sqrt{3}km$.

D) In $\Delta ABD$, $\sin 30°=\frac{AB}{BD}$

$\Rightarrow \frac{1}{2}=\frac{\frac{3}{2}}{BD}$

$\Rightarrow BD=\frac{3}{2}\times 2$

$\Rightarrow BD=3$

$\therefore$ Distance between  $B\text{\hspace{0.17em}and\hspace{0.17em}}D$ is  $3km$.