A straight road leads to the foot of a tower $200 \; m$ high. From the top of the tower, the angles of depression of two cars standing on the road are observed to be $45^\circ$ and $60^\circ$ respectively. Find the distance between the two cars. Give your answer correct to two decimal places. Take $\sqrt{3} = 1.732$.
$OT$: Tower of height $200 \; m$
$C_1, \; C_2$: Positions of the two cars
$C_1C_2$: Distance between the two cars
In $\triangle OTC_1$,
$\dfrac{OT}{OC_1} = \tan 60^\circ = \sqrt{3}$
i.e. $\;$ $OC_1 = \dfrac{OT}{\sqrt{3}} = \dfrac{200}{\sqrt{3}} \; m$ $\;\;\; \cdots \; (1)$
In $\triangle OTC_2$,
$\dfrac{OT}{OC_2} = \tan 45^\circ = 1$
i.e. $\;$ $OC_2 = OT = 200 \; m$ $\;\;\; \cdots \; (2)$
Now, $\;$ $C_1 C_2 = OC_2 - OC1$
i.e. $\;$ $C_1 C_2 = 200 - \dfrac{200}{\sqrt{3}}$ $\;\;\;$ [in view of equations $(1)$ and $(2)$]
i.e. $\;$ $C_1 C_2 = 200 \left(\dfrac{\sqrt{3} - 1}{\sqrt{3}}\right) = 84.53 \; m$ $\;\;\;$ [correct to two decimal places]