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Answer :
Let AB be the tower of height h metres and let C be a point at a distance of 30 m from the foot of the tower. The angle of elevation of the top of the tower from point C is given as 30º.
In \( ∆ \ CAB \), we have
\( \frac{AB}{CA} \ = \ tan30º \)
\( \frac{h}{30} \ = \ \frac{1}{ \sqrt{3}} \ => \ h \ = \ \frac{30}{ \sqrt{3}} \ = \ 10 \sqrt{3} \)
Hence, the height of the tower is \( 10 \sqrt{3} \) metres.