Question : The angle of elevation of the top of a tower $25 \sqrt{3} \mathrm{~m}$ high from two points on the level ground on its opposite sides are 45° and 60°. What is the distance (in m) between the two points (correct to one decimal place)?
Option 1: 45.3
Option 2: 58.4
Option 3: 68.3
Option 4: 50.6
Correct Answer: 68.3
Solution :
Let the distances from the two points on the ground to the base of the tower as $d_1$ and $d_2$.
The angles of elevation from these points to the top of the tower are given as 45° and 60°, respectively.
In triangle ABD,
$\tan(45^\circ) = \frac{AD}{BD}$
$⇒\tan(45^\circ) = \frac{25\sqrt{3}}{d_1}$
$⇒1 = \frac{25\sqrt{3}}{d_1}$
$⇒d_1 = 25\sqrt{3}= 25 ×1.732=43.3 \;\text{m}$
In triangle ADC,
$\tan(60^\circ) = \frac{AD}{DC}$
$⇒\tan(60^\circ) = \frac{25\sqrt{3}}{d_2}$
$⇒\sqrt{3} = \frac{25\sqrt{3}}{d_2}$
$⇒d_2 = 25 \;\text{m}$
Therefore, the distance between the two points $=d_1 + d_2 = 43.3 + 25 = 68.3 \;\text{m}$
Hence, the correct answer is 68.3 m.
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