From the top of a 25 m high cliff,the angle | vedclass

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(C) Let $AB$ be the cliff of height $25 \ m$ and $CD$ be the tower.Let the angle of elevation of the top of the tower $D$ from the top of the cliff $B

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(C) Let $AB$ be the cliff of height $25 \ m$ and $CD$ be the tower.Let the angle of elevation of the top of the tower $D$ from the top of the cliff $B$ be $\alpha$.Let the angle of depression of the foot of the tower $C$ from the top of the cliff $B$ be $\alpha$.Draw $BE \perp CD$,where $E$ is on $CD$.In $	riangle ABC$,$	an \alpha = \frac{AB}{AC} = \frac{25}{AC}$.In $	riangle BDE$,$	an \alpha = \frac{DE}{BE}$.Since $BE = AC$,we have $\frac{DE}{BE} = \frac{25}{BE}$,which implies $DE = 25 \ m$.Since $CE = AB = 25 \ m$,the total height of the tower $CD = CE + DE = 25 + 25 = 50 \ m$.

From the top of a $25 \ m$ high cliff,the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find the height of the tower in $m$.

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