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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