Question : A pole of length 7 m is fixed vertically on the top of a tower. The angle of elevation of the top of the pole observed from a point on the ground is 60° and the angle of depression of the same point on the ground from the top of the tower is 45°. The height (in m) of the tower is:
Option 1: $7(2 \sqrt{3}-1)$
Option 2: $\frac{7}{2}(\sqrt{3}+2)$
Option 3: $7 \sqrt{3}$
Option 4: $\frac{7}{2}(\sqrt{3}+1)$
Correct Answer: $\frac{7}{2}(\sqrt{3}+1)$
Solution :
Let pole be AD and tower be BD.
Given,
$AD = 7\ m$
In $\triangle BCD$
$\tan45^\circ = \frac{BD}{BC}$
⇒ $1 = \frac{BD}{BC}$
⇒ $BC = BD$
In $\triangle ABC$
$\tan60^\circ = \frac{AB}{BC}$
⇒ $\sqrt3 = \frac{AB}{BC}$
⇒ $AB = \sqrt3BC$
⇒ $AD + BD = \sqrt3 BD$
⇒ $\sqrt3BD - BD = 7$
⇒ $BD(\sqrt3 - 1) = 7$
⇒ $BD= \frac{7}{(\sqrt3 - 1)} × \frac{(\sqrt3 + 1)}{(\sqrt3 + 1)}$
$= \frac{7(\sqrt3 + 1)}{(\sqrt3)^2 - 1^2}$
$= \frac{7(\sqrt3 + 1)}{(3 - 1)}$
$= \frac{7}{2}(\sqrt3 + 1)$
Hence, the correct answer is $\frac{7}{2}(\sqrt3 + 1)$.
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