Observing from the top of a building,the angl | vedclass

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(D) Let the height of the building be $AB = 36 \, m$ and the height of the tower be $CD = h$. Let the distance between the building and the tower be $

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(D) Let the height of the building be $AB = 36 \, m$ and the height of the tower be $CD = h$. Let the distance between the building and the tower be $x$.From the top of the building $(A)$,the angle of elevation to the top of the tower $(D)$ is $30^{\circ}$.In the right-angled triangle formed by the horizontal line from $A$ to the tower at point $E$,we have $	an(30^{\circ}) = \frac{DE}{AE} = \frac{h - 36}{x}$.Since $	an(30^{\circ}) = \frac{1}{\sqrt{3}}$,we get $x = \sqrt{3}(h - 36)$.From the top of the building $(A)$,the angle of depression to the base of the tower $(C)$ is $60^{\circ}$.In the right-angled triangle $ABC$,$	an(60^{\circ}) = \frac{AB}{BC} = \frac{36}{x}$.Since $	an(60^{\circ}) = \sqrt{3}$,we get $x = \frac{36}{\sqrt{3}} = 12\sqrt{3}$.Equating the two expressions for $x$: $\sqrt{3}(h - 36) = 12\sqrt{3}$.Dividing by $\sqrt{3}$,we get $h - 36 = 12$,which implies $h = 48 \, m$.

Observing from the top of a building,the angle of elevation of the top of a tower is $30^{\circ}$ and the angle of depression of the base of the tower is $60^{\circ}$. If the height of the building is $36 \, m$,find the height of the tower (in $m$).

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