If the angles of elevation of the top of a tower from the top and front of a pole of height 15 m are $$30^\circ$$ and $$60^\circ$$ respectively, then the height of the tower(in meters) is

Solution

As per the given question,

Let the height of the tower is h.

Now, in the $$\triangle$$ECD,

$$\tan 30=\dfrac{CD}{EC}---------------(i)$$

In the $$\triangle$$ABD

$$\Rightarrow \tan 60=\dfrac{BD}{AB}$$

$$\Rightarrow \tan 60=\dfrac{15+CD}{AB}--------------(ii)$$

AB=EC

Hence from equation (i) and (ii)

$$\Rightarrow \dfrac{\dfrac{CD}{EC}}{\dfrac{15+CD}{EC}}=\dfrac{\tan 30}{\tan 60}$$

$$\Rightarrow \dfrac{CD}{15+CD}=\dfrac{1}{3}$$

$$\Rightarrow 3CD=15+CD$$

$$\Rightarrow 2CD=15$$

$$\Rightarrow CD=\dfrac{15}{2}$$

Hence the height of the tower$$ =15+\dfrac{15}{2}=\dfrac{45}{2}=22\dfrac{1}{2}$$