Question

If the angles of the depression and elevation of the top of a tower of height h from the top and bottom of a second tower are x and y respectively then the height of the second tower is:
(a) h (cot y + cot x)
(b) h (tan x + tan y)
(c) h (1 + tan x cot y)
(d) h (tan y cot x + 1)

Answer 1

Hint:First of all, draw a figure in which there are two towers AB and CD of height h and $$h_1$$ respectively. Now, from the top of the tower CD, draw an angle of depression x and from the bottom of tower CD, draw the angle of elevation y at the top of the tower AB. Now, use tan x and tan y to find the value of $$h_1$$ that is the height of the second tower.

Complete step-by-step answer:
In this question, we are given that the angles of depression and elevation of the top of a tower of height h from the top and bottom of a second tower are x and y respectively then we have to find the height of the second tower. Let us assume two towers of height h and $$h_1$$.We are given the angles of depression of the top of the tower of height h from the top of the second tower is x. Also, the angle of elevation of the top of the tower of height h from the bottom of the second tower is y. So, we get,

We know that DE is parallel to FA and $$\mathrm{\angle }EDF\text{ and }\mathrm{\angle }DAF$$ are alternate angles. So,
$$\mathrm{\angle }EDF=\mathrm{\angle }DAF=x$$
We know that,
$$\tan \theta ={\displaystyle \frac{perpendicular}{base}}$$
Let us consider a triangle ABC, so we get,
$$\tan C={\displaystyle \frac{P}{B}}={\displaystyle \frac{AB}{BC}}$$
By substituting the value of $$\mathrm{\angle }C$$ as y and AB = h, we get,
$$\tan y={\displaystyle \frac{h}{BC}}$$
$$BC={\displaystyle \frac{h}{\tan y}}.....\left(i\right)$$
Let us consider triangle AFD, we get,
$$\tan A={\displaystyle \frac{P}{B}}={\displaystyle \frac{DF}{AF}}$$
By substituting the values of $$\mathrm{\angle }A$$ = x and FA = BC, we get,
$$\tan x={\displaystyle \frac{DF}{BC}}....\left(ii\right)$$
We can write DF as DC – FC. We know that FC = AB = h and $$DC=h_1$$. So, we get,
$$DF=h_1-h$$
By substituting the value of AF in equation (ii), we get,
$$\tan x={\displaystyle \frac{h_1-h}{BC}}$$
$$BC={\displaystyle \frac{(h_1-h)}{\tan x}}....\left(iii\right)$$
Now, by equating the value of BC from equation (i) and (iii), we get,
$${\displaystyle \frac{h}{\tan y}}={\displaystyle \frac{h_1-h}{\tan x}}$$
$${\displaystyle \frac{h}{\tan y}}+{\displaystyle \frac{h}{\tan x}}={\displaystyle \frac{h_1}{\tan x}}$$
$$h_1={\displaystyle \frac{h\tan x}{\tan y}}+{\displaystyle \frac{h\tan x}{\tan x}}$$
By canceling the like terms and using $${\displaystyle \frac{1}{\tan \theta }}=\cot \theta$$, we get,
$$h_1=h\tan x\cot y+h$$
$$h_1=h(1+\tan x\cot y)$$
So, we get the height of the second tower as h (1 + tan x cot y).
Hence, the option (c) is the right answer.

Note: In this type of questions, the most important point is to make the proper diagram to visualize the question. In this question, many students make this mistake of making the angle of depression and elevation from the top of tower 1 to the top and bottom of tower 2 by taking $$h>h_1$$. But actually, we need to do the reverse and take $$h_1>h$$, because then only we can make the angle of depression from the top of the second tower to the top of the first tower. So, all these points must be taken care of.