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$P$ and $Q$ are two points observed from the top of a building $10 \sqrt{3} m$ high if the angles of depression of the points are complementary and $PQ = 20 m$ , then the distance of $P$ from the building is: (P being the farther point)

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30 m

75 m

45 m

40 m

answer is A.

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Detailed Solution

P

and

Q

are two points observed from the top of a building

10 \sqrt{3} m

high if the angles of depression of the points are complementary and

PQ = 20 m

We have to find the distance of

P

from the building.
We use

tanθ = \frac{height}{base}

.
It is given that the Angle of depressions are

x

and

90 - x

as shown in the figure.
From,

△ POR

tanx = \frac{OR}{PR}

\Rightarrow tanx = \frac{h}{PR}

⟹ PR = \frac{h}{tanx}

......(1)
From

△ ROQ

\tan (90^{\circ} - x) = \frac{OR}{QR}

\Rightarrow \tan (90^{\circ} - x) = \frac{h}{QR}

⟹ QR = \frac{h}{\tan (90^{\circ} - x)}

⟹ QR = \frac{h}{cotx}

[∵ \tan (90^{\circ} - θ) = cotθ]

⟹ QR = htanx

.....(2)

[∵ \frac{1}{cotx} = tanx]

∵ PQ = PR - QR

⟹ 20 = \frac{h}{tanx} - htanx

⟹ 20 tanx = h - h \tan^{2} x

⟹ 20 tanx = 10 \sqrt{3} - 10 \sqrt{3} \tan^{2} x

[∵ h = 10 \sqrt{3}]

⟹ 10 \sqrt{3} \tan^{2} x + 20 tanx - 10 \sqrt{3} = 0

Dividing all by

10 \sqrt{3}

we get,

⟹ \tan^{2} x + \frac{2}{\sqrt{3}} tanx - 1 = 0

⟹ tanx = \frac{- \frac{2}{\sqrt{3}} \pm \sqrt{{(\frac{2}{\sqrt{3}})}^{2} + 4}}{2}

(Using quadratic formula)

⟹ tanx = \frac{- \frac{2}{\sqrt{3}} \pm \frac{4}{\sqrt{3}}}{2}

⟹ tanx = - \frac{1}{\sqrt{3}} \pm \frac{2}{\sqrt{3}}

⟹ tanx = \frac{1}{\sqrt{3}}

tanx = - \sqrt{3}

Taking only,

tanx = \frac{1}{\sqrt{3}}

as it is not possible to take negative value.

⟹ x = 30^{\circ}

[

∵

tan

30^{\circ}

\frac{1}{\sqrt{3}}

]
From (1),

PR = \frac{10 \sqrt{3}}{tanx}

\Rightarrow PR = 10 \sqrt{3} \times \sqrt{3}

\Rightarrow PR = 30 m

Hence, distance of point P is

30 m

.
Therefore, the correct option is 1.

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