Question 121
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=ECHence 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$$ =15+\dfrac{15}{2}=\dfrac{45}{2}=22\dfrac{1}{2}$$
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