In Fig. $\displaystyle 3$, AB is tower of height $\displaystyle 50$ m. A man standing on its top, observes two cars on the opposite sides of the tower with angles of depression $\displaystyle 30$° and $\displaystyle 45$° respectively. Find the distance between the two cars. Fig.

\[\begin{array}{c} \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{A B}{B C} \\ \Rightarrow B C=A B \sqrt{3} m \ldots \ldots . .(i) \\ \tan 45^{\circ}=\frac{A B}{B D}=1 \\ \Rightarrow B D=A B \ldots \ldots .(i i) \\ \text { (i) and }(i i) B C+B D=A B(\sqrt{3}+1) \\ \Rightarrow C D=50(\sqrt{3}+1) \mathrm{m} \end{array} \]The mean of the following frequency distribution is $\displaystyle 25$ . Find the value of f.Class : & $\displaystyle 0$-$\displaystyle 10$ & $\displaystyle 10$-$\displaystyle 20$ & $\displaystyle 20$-$\displaystyle 30$ & $\displaystyle 30$-$\displaystyle 40$ & $\displaystyle 40$-$\displaystyle 50$

Class	$\displaystyle \boldsymbol{x}$	$\displaystyle f$	$\displaystyle f x$
$\displaystyle 0$-$\displaystyle 10$	$\displaystyle 5$	$\displaystyle 5$	$\displaystyle 25$
$\displaystyle 10$-$\displaystyle 20$	$\displaystyle 15$	$\displaystyle 18$	$\displaystyle 270$
$\displaystyle 20$-$\displaystyle 30$	$\displaystyle 25$	$\displaystyle 15$	$\displaystyle 375$
$\displaystyle 30$-$\displaystyle 40$	$\displaystyle 35$	$\displaystyle f$	$\displaystyle 35 f$
$\displaystyle 40$-$\displaystyle 50$	$\displaystyle 45$	$\displaystyle 6$	$\displaystyle 270$
		$\displaystyle 44+f$	$\displaystyle 940+35 f$

\[\begin{array}{c} \bar{x}=25=\frac{940+35 f}{44+f} \\ \Rightarrow f=16 \end{array} \]Or Find the mean of the following data using assumed mean method :

Class :	$\displaystyle 0$-$\displaystyle 5$	$\displaystyle 5$-$\displaystyle 10$	$\displaystyle 10$-$\displaystyle 15$	$\displaystyle 15$-$\displaystyle 20$	$\displaystyle 20$-$\displaystyle 25$
Frequency :	$\displaystyle 8$	$\displaystyle 7$	$\displaystyle 10$	$\displaystyle 13$	$\displaystyle 12$

Class	$\displaystyle \boldsymbol{x}$	$\displaystyle f$	$\displaystyle d=x-12 \cdot 5$	$\displaystyle f d$
$\displaystyle 0$-$\displaystyle 5$	$\displaystyle 2 \cdot 5$	$\displaystyle 8$	-$\displaystyle 10$	-$\displaystyle 80$
$\displaystyle 5$-$\displaystyle 10$	$\displaystyle 7.5$	$\displaystyle 7$	-$\displaystyle 5$	-$\displaystyle 35$
$\displaystyle 10$-$\displaystyle 15$	$\displaystyle 12 \cdot 5$	$\displaystyle 10$	$\displaystyle 0$	$\displaystyle 0$
$\displaystyle 15$-$\displaystyle 20$	$\displaystyle 17.5$	$\displaystyle 13$	$\displaystyle 5$	$\displaystyle 65$
$\displaystyle 20$-$\displaystyle 25$	$\displaystyle 22 \cdot 5$	$\displaystyle 12$	$\displaystyle 10$	$\displaystyle 120$
		$\displaystyle 50$		$\displaystyle 70$

$\displaystyle \begin{array}{c}\bar{x}=12 \cdot 5+\frac{70}{50} =13 \cdot 9\end{array}$