An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:
21.6 m
23.2 m
24.72 m
None of these
Explanation:
Let AB be the observer and CD be the tower.
Draw BE $\bot$ CD.
Then, CE = AB = 1.6 m,
BE = AC = 20$\sqrt{3}$ m.
$\dfrac{DE}{BE}$ = tan 30°=1/$\sqrt{3}$
$\Rightarrow \dfrac{20\sqrt{3}}{\sqrt{3}}$m=20 m
$\therefore$ CD = CE + DE = (1.6 + 20) m = 21.6 m.
Let AB be the observer and CD be the tower.
Draw BE $\bot$ CD.
Then, CE = AB = 1.6 m,
BE = AC = 20$\sqrt{3}$ m.
$\dfrac{DE}{BE}$ = tan 30°=1/$\sqrt{3}$
$\Rightarrow \dfrac{20\sqrt{3}}{\sqrt{3}}$m=20 m
$\therefore$ CD = CE + DE = (1.6 + 20) m = 21.6 m.