# Q. From the top of a 7m high building the angle of elevation of the top of a cable tower is 60$°$ and the angle of depression of the foot of the tower is 30$°$. Find the height of the tower .

Dear Student,

Let AB be the building and CD be the tower such that <EAD = 60o and <EAC = <ACB = 30o

Now, In triangle ABC, tan 30 = $\frac{1}{\sqrt{3}}$ = AB/BC

So, BC = AE = 7$\sqrt{3}$m

Again in triangle AED,

tan 60 = $\sqrt{3}$ = DE/AE

So, DE= AE.$\sqrt{3}$ = 7$\sqrt{3}×\sqrt{3}$ m = 21 m

Height of the cable tower = h + 7 = 21 + 7 m  = 28 m

Regards

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