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(D) Let the height of the building $AB$ be $H$ metres.Let $Q$ be the top of the tower and $P$ be the base of the tower. Given $QP = 30 \ m$.In $\Delta

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$12.68$

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(D) Let the height of the building $AB$ be $H$ metres.Let $Q$ be the top of the tower and $P$ be the base of the tower. Given $QP = 30 \ m$.In $\Delta APQ$,the angle of elevation from $A$ to $Q$ is equal to the angle of depression from $Q$ to $A$,which is $60^{\circ}$.$	an 60^{\circ} = \frac{QP}{AP} = \frac{30}{AP} = \sqrt{3}$$AP = \frac{30}{\sqrt{3}} = 10 \sqrt{3} \ m$.Let $R$ be a point on $QP$ such that $BR$ is horizontal. Then $BR = AP = 10 \sqrt{3} \ m$.The angle of elevation from $B$ to $Q$ is equal to the angle of depression from $Q$ to $B$,which is $45^{\circ}$.In $\Delta BRQ$,$	an 45^{\circ} = \frac{QR}{BR} = 1$.$QR = BR = 10 \sqrt{3} \ m$.The height of the building $H = AB = RP = QP - QR$.$H = 30 - 10 \sqrt{3} = 30 - 10(1.732) = 30 - 17.32 = 12.68 \ m$.

$A$ tower is $30 \ m$ high. An observer from the top of the tower makes an angle of depression of $60^{\circ}$ at the base of the building and an angle of depression of $45^{\circ}$ at the top of the building. What is the height of the building in metres?

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