#

Q. From the top of a 7m high building the angle of elevation of the top of a cable tower is 60$\xb0$ and the angle of depression of the foot of the tower is 30$\xb0$. Find the height of the tower .

Here is the solution of your asked query:

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

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$\sqrt{3}\times \sqrt{3}$ m = 21 m

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

Regards

**
**