An observer finds that the angular elevation of a tower is θ. On advancing 3m towards the tower the elevation is 45∘and on advancing 2m nearer, the elevation is (90∘-θ). Then what is the height of the tower?

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An observer finds that the angular elevation of a tower is $θ$ . On advancing $3$ m towards the tower the elevation is $45^{\circ}$ and on advancing $2$ m nearer, the elevation is $(90^{\circ} - θ)$ . Then what is the height of the tower?

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2

4

6

8

answer is C.

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Detailed Solution

Given that, angular elevation of a tower is

θ

and on advancing

3

m towards the tower the elevation is

45^{\circ}

and on advancing

2

m nearer, the elevation is

(90^{\circ} - θ)

∠ PAQ = θ

∠ PBQ = 45^{o}

∠ PCQ = 90^{o} - θ

AB = 3

BC = 2

We have the formula,

tanθ = \frac{Height}{distance}

cotθ = \frac{distance}{Height}

We have to find the height,
Taking

A

as the initial point of observing tower

PQ

B

= 3m away from

A

and

C

= 2m away from

B

.
In triangle PQC,
tan

(90^{o} - θ)

\frac{PQ}{QC}

\Rightarrow

cot

θ

\frac{h}{QC}

…..(1)

\Rightarrow h =

QC cot

θ

In triangle PBQ,
tan

(45^{o})

\frac{h}{2 + QC}

\Rightarrow \frac{h}{2 + QC} = 1

\Rightarrow h - 2 = QC

……(2)
In triangle PAQ,
tan

θ

\frac{h}{5 + QC}

\Rightarrow \frac{h}{5 + QC} = tanθ

……(3)

\Rightarrow h = tanθ (5 + QC)

Multiplying (1) and (3)
cot

θ \times

tan

θ

\frac{h}{QC} \times \frac{h}{5 + QC}

\Rightarrow 1 = \frac{h^{2}}{5 QC + {QC}^{2}}

\Rightarrow 5 QC + {QC}^{2} = h^{2}

From (2),

\Rightarrow 5 (h - 2) + {(h - 2)}^{2} = h^{2}

\Rightarrow 5 h - 10 + h^{2} + 4 - 4 h = h^{2}

\Rightarrow h^{2} + h - 6 = h^{2}

\Rightarrow h = 6

m
Thus, the height is 6m.
Therefore the correct option is 3.

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