Question : Two points A and B are on the ground and on opposite sides of a tower. A is closer to the foot of the tower by 42 m than B. If the angles of elevation of the top of the tower, as observed from A and B are 60° and 45°, respectively, then the height of the tower is closest to:
Option 1: 87.6 m
Option 2: 98.6 m
Option 3: 88.2 m
Option 4: 99.4 m
Correct Answer: 99.4 m
Solution :
PQ is a tower and let AQ be $x$ m and BQ = $x + 42$
In $\triangle APQ$
$\tan 60° = \frac{PQ}{AQ}$
⇒ $\sqrt3 = \frac{PQ}{x}$
⇒ $PQ = \sqrt3 x$.....................................(1)
In $\triangle PBQ$
$\tan 45° = \frac{PQ}{QB}$
⇒ $1 = \frac{PQ}{x + 42}$
⇒ $PQ = (x + 42)$......................................(2)
From equation (1) and equation (2)
⇒ $\sqrt3x = x + 42$
⇒ $x (\sqrt3 - 1) = 42$
⇒ $x = \frac{42}{(\sqrt3 - 1)}$
⇒ $x = \frac{42}{(\sqrt3 - 1)} \times \frac{(\sqrt3 + 1)}{(\sqrt3 + 1)}$
= $\frac{42 (1.732 + 1)}{(3 - 1)}$
= $21 \times 2.732 \approx 57.4$
From equation (2)
PQ = 57.4 + 42 = 99.4 m
Hence, the correct answer is 99.4 m.
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