Question
The top of a hill when observed from the top and bottom of a building of height $$h$$ is at angles of elevation $$p$$ and $$q$$ respectively. What is the height of the hill ?
A.
$$\frac{{h\cot q}}{{\cot q - \cot p}}$$
B.
$$\frac{{h\cot p}}{{\cot p - \cot q}}$$
C.
$$\frac{{2h\tan p}}{{\tan p - \tan q}}$$
D.
$$\frac{{2h\tan q}}{{\tan q - \tan p}}$$
Answer :
$$\frac{{h\cot p}}{{\cot p - \cot q}}$$
Solution :
Let height of hill $$= H$$
& horizontal distance between building & hill $$= d$$
$$\eqalign{ & \tan q = \frac{H}{d} \cr & \Rightarrow d = \frac{H}{{\tan q}} = H\cot q \cr & \tan p = \frac{{\left( {H - h} \right)}}{d} \cr & \Rightarrow d = \left( {H - h} \right)\cot p \cr & \Rightarrow H\cot q = \left( {H - h} \right)\cot p \cr & H = \frac{{h\cot p}}{{\cot p - \cot q}} \cr} $$
Let height of hill $$= H$$
& horizontal distance between building & hill $$= d$$
$$\eqalign{ & \tan q = \frac{H}{d} \cr & \Rightarrow d = \frac{H}{{\tan q}} = H\cot q \cr & \tan p = \frac{{\left( {H - h} \right)}}{d} \cr & \Rightarrow d = \left( {H - h} \right)\cot p \cr & \Rightarrow H\cot q = \left( {H - h} \right)\cot p \cr & H = \frac{{h\cot p}}{{\cot p - \cot q}} \cr} $$