To a man standing outside his house, the angles of elevation of the top and bottom of a window are $60^\circ$ and $45^\circ$ respectively. If the height of the man is $180 \; cm$ and if he is $5 \; m$ away from the wall, what is the height of the window and the wall? [Given: $\sqrt{3} = 1.732$]
$W_1$: Top of window
$W_2$: Bottom of window
$OW_2$: Wall
$AB$: Man of height $180 \; cm = 1.8 \; m$
$OA$: Distance of man from wall $= 5 \; m$
Draw $BB' \perp O W_2 W_1$
In the figure, $\;$ $AB = OB' = 1.8 \; m$ $\;$ and $\;$ $BB' = OA = 5 \; m$
In $\triangle W_2B'B$, $\;$ $\dfrac{W_2B'}{BB'} = \tan 45^\circ = 1$
$\implies$ $W_2 B' = BB' = 5 \; m$
In $\triangle W_1B'B$, $\;$ $\dfrac{W_1B'}{BB'} = \tan 60^\circ = \sqrt{3}$
$\implies$ $W_1B' = BB' \sqrt{3} = 5 \sqrt{3} \; m$
$\therefore \;$ Height of window $= W_1W_2 = W_1 B' - W_2 B' = 5 \left(\sqrt{3} - 1\right) = 5 \times 0.732 = 3.66 \; m$
Height of wall $= W_2 O = W_2 B' + B'O = 5 + 1.8 = 6.8 \; m$