A pole of length 7 m is fixed vertically on the top of a tower. The angle of elevation of the top of the pole observed from a point on the ground is $$60^\circ$$ and the angle of depression of the same point on the ground from the top of the tower is $$45^\circ$$. The height (in m) of the tower is:

As per the given question,

Length of the pole is DC=7m
The angle of elevation of the top of the pole to ground$$=60^\circ$$

The angle of elevation of the top of the tower $$=45^\circ$$

Now, $$\tan 60^\circ=\dfrac{BC}{AB}$$

$$\Rightarrow \sqrt{3}=\dfrac{BC}{AB}$$

$$\Rightarrow AB=\dfrac{BC}{\sqrt{3}}------(i)$$

Now, In the $$\triangle ABD$$

$$\Rightarrow \tan 45^\circ=\dfrac{BD}{AB}$$

$$\Rightarrow 1=\dfrac{BD}{AB}$$

$$\Rightarrow AB=BD------(ii)$$

From (i) and (ii)

$$\Rightarrow BD=\dfrac{DC+BD}{\sqrt{3}}$$

$$\Rightarrow BD=\dfrac{7+BD}{\sqrt{3}}$$

$$\Rightarrow BD\sqrt{3}=7+BD$$

$$\Rightarrow BD(\sqrt{3}-1)=7$$

$$\Rightarrow BD=\dfrac{7}{\sqrt{3}-1}$$

$$\Rightarrow BD=\dfrac{7(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$$

$$\Rightarrow BD=\dfrac{7(\sqrt{3}+1)}{3-1}$$

$$\Rightarrow BD=\dfrac{7(\sqrt{3}+1)}{2}$$