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The angles of elevation of the top of a building from the top and bottom of a tree are $x$ and $y$ respectively. If the height of the tree is $h$ m, then, in meter, the height of the building is:

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\frac{hcotx}{cotx + coty}

\frac{hcoty}{cotx + coty}

\frac{hcotx}{cotx - coty}

\frac{hcoty}{cotx - coty}

answer is A.

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Detailed Solution

It is given that the angles of elevation of the top of a building from the top and bottom of a tree are

x

and

y

.
The height of the tree is

h

m.
We have to find the height of the building.
We use

tanθ = \frac{perpendicular}{base} .

△ AED

tanx = \frac{DE}{AE}

\Rightarrow tanx = \frac{H - h}{AE}

------- (1)

\Rightarrow AE = \frac{H - h}{tanx}

------(A)
In

△ DBC

tany = \frac{DC}{BC}

\Rightarrow tany = \frac{H}{AE}

------- (2) [since BC=AE]

\Rightarrow AE = \frac{H}{tany}

------ (B)
Now on comparing

A

and ,

\frac{H - h}{tanx} = \frac{H}{tany}

\Rightarrow (H - h) tany = H . tanx

\Rightarrow H tany - H tanx = h tany

\Rightarrow H = \frac{htany}{tany - tanx}

\Rightarrow

= \frac{\frac{h}{coty}}{\frac{1}{coty} - \frac{1}{cotx}}

\Rightarrow H = \frac{\frac{\frac{h}{coty}}{cotx - coty}}{cotx . coty}

\Rightarrow H = \frac{hcotx}{cotx - coty}

Hence, the height of the building is

\frac{hcotx}{cotx - coty} m

.
Therefore, the correct option is 1

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