In the given diagram, AB is a vertical tower 100 m away from the foot of a 30 storied building CD. The angles of depression from the point C and E, (E being the mid-point of CD), are 35° and 14° respectively. (Use mathematical table for the required values rounded off correct to two places of decimals only.)
Find the height of the:
(a) tower AB
(b) building CD
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From figure,
Let height of the building CD be H meters and height of tower AB be h meters.
Given,
BD = 100 m
E is the midpoint of CD
∴ CE = $\frac{CD}{2} = \frac{H}{2}$
From figure,
⇒ ∠CAP = ∠XCA = 35° [Alternate angles are equal]
⇒ ∠EAP = ∠YEA = 14° [Alternate angles are equal]
From figure,
PD = AB = h
AP = BD = 100 m
In ΔACP,
$$\begin{aligned}\Rightarrow \text{tan 35°} = \frac{\text{Perpendicular}}{\text{Base}}\\[1em] \Rightarrow \text{tan 35°} = \frac{CP}{AP} \\[1em] \Rightarrow \text{tan 35°} = \frac{CD - PD}{AP} \\[1em] \Rightarrow 0.70 = \frac{H - h}{100} \\[1em] \Rightarrow 0.70 × 100 = H - h \\[1em] \Rightarrow H - h = 70 .....(1)\end{aligned}$$
In ΔAEP,
$$\begin{aligned}\Rightarrow \text{tan 14°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 14°} = \frac{EP}{AP}\\[1em] \Rightarrow \text{tan 14°} = \frac{ED - PD}{AP} \\[1em] \Rightarrow 0.25 = \frac{\frac{H}{2} - h}{100} \\[1em] \Rightarrow \frac{H}{2} - h = 0.25 × 100 \\[1em] \Rightarrow \frac{H}{2} - h = 25 .....(2)\end{aligned}$$
Subtracting equation (2) from (1), we get:
$$\begin{aligned}\Rightarrow H - h - \Big({\frac{H}{2} - h}\Big) = 70 - 25 \\[1em] \Rightarrow H - h - {\frac{H}{2} + h} = 45 \\[1em] \Rightarrow H - \frac{H}{2} = 45 \\[1em] \Rightarrow \frac{H}{2} = 45 \\[1em] \Rightarrow H = 45 \times 2 = 90 \text{ m}.\end{aligned}$$
⇒ H - h = 70
⇒ 90 - h = 70
⇒ h = 90 - 70
⇒ h = 20 m.
Hence, height of the tower AB = 20 m and height of the building CD = 90 m.