Question

$\huge\mathbb{Question}$


p and q are two point observed from the top of a building 10√3 m hight. If the angle of depression of the point are complementary and pq = 20m , then the distance of p from the building is?​

Answer 1

Step-by-step explanation:

cot θ = 1. ⇒ l = 10 m. ∴

The distance of P from the building = 10 + 20 = 30 m.

This discussion on P and Q are two points observed from the top of a building 10√3 m high.

Answer 2

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♠ Correct Question :-

P and Q are two point observed from the top of a building 10√3 m height . If the angle of depression of the point are complementry and PQ = 20m , then the distance of P from the buliding is (Given P is farther point)

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♠ Answer :-

• Distance of point P from building is 30m.

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♠ Solution :-

As the angles are complementary So their sum must be 90° .

$\large \bigstar \: \underline{ \pmb{ \mathfrak{ \text{A}ccording \: to \: \text{A}ttached \: figure }}}$

$\Large \pmb{In \: \triangle \: SRQ \: :-}$

$\sf \tan(90 - x) = \dfrac{10 \sqrt{3} }{y} \\ \\ \bf :\longmapsto cot \: x = \frac{10 \sqrt{3} }{y} \: \:\: - - - (i)$

Now ,

$\Large \pmb{In \: \triangle \: SRP \: :-}$

$\bf tan \: x = \dfrac{10 \sqrt{3} }{y + 20} \: \: \: \: - - - - (ii)$

Mulyiplying (i) and (ii) We get,

$\sf \tan x. \cot x = \dfrac{10 \sqrt{3}}{y} \times \frac{10 \sqrt{3} }{20 + y} \\ \\ :\longmapsto \sf 1 = \frac{300}{y(y + 20)} \\ \\:\longmapsto \sf1 = \frac{300}{y {}^{2} + 20y} \\ \\ \bf:\longmapsto y {}^{2} + 20y - 300 = 0 \\ \\ :\longmapsto \sf y {}^{2} - 10y + 30y - 300 = 0 \\ \\:\longmapsto \sf y(y - 10) + 30(y - 10) = 0 \\ \\ :\longmapsto\sf (y - 10)(y + 30) = 0 \\ \\ \large \sf \purple{ :\longmapsto \underline {\boxed{{\bf y = 10 \: \: or \: \: y = - 30 } }}}$

As y is distance So it can't be Negative

Hence,

$\Large\purple{ :\longmapsto \underline {\boxed{{\bf y = 10m} }}}$

So,

• Distance of point P from building = 10 + 20

$\\ \\ \pink{ \LARGE:\longmapsto \mathfrak{Distance=30m}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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