Question

Topic - Height And Distances
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P and Q are two point observed from the top of a building 10√3 m hight . if the angleof depression of the point are complementry and PQ = 20m. Find the distance of P from the buliding .

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Answer 1

$\large \bf \clubs \: Question :-$

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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$\large \bf \clubs \:  Given :-$

• P and Q are two point observed from the top of a building

• Height of Building 10√3 m .

• The angle of depression of the point are complementry

• Distance PQ = 20m

• P is father Point

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$\large \bf \clubs \:  To \: Find :-$

• The Distance of the P from Building .

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$\large \bf \clubs \: Answer :-$

Distance of point P from building is 30m.

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$\large \bf \clubs \:Solution:-$

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

According to the Attached Figure:

$\large \bf In \: \triangle \: SRQ$

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

Now ,

$\bf \large In \: \triangle \: SRP$

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

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 meters

That is

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

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Answer 2

Given :

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• P and Q are two points observed from the top of a building 10√3 m high. if the angle of depression of the point are complementary and PQ = 20m.

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To Find :

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• Distance of P from the building?

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Solution :

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• Let the building be RS and ∠P be x. Also, let distance b/w building and point Q be m meter. So, distance b/w building and point P is (m + 20) meter. We know that ∠P + ∠Q = 90° (It is given in the Question that angles are complementary). Therefore, ∠Q = 90° - x.

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⚝ Things to know before solving :

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• $\begin{cases} & \bf{\red{tan\:\theta = \dfrac{Perpendicular}{Base}}} \\ \\ \\ & \bf{\orange{tan(90^{\circ} - \theta) = cot\:\theta}} \\ \\ \\ & \bf{\green{tan\:\theta\:.\:cot\:\theta = 1}} \end{cases}$

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• So, let's solve it! $\red{\clubsuit}$

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⚝ In ∆SRQ :

$\\ :\implies\:\sf tan(90^{\circ} - x) = \dfrac{SR}{RQ}$

$\\ :\implies\:\bf\pink{cot\:x = \dfrac{10\sqrt{3}}{m}} \:\dashrightarrow\:\blue{Eq^{n}\:(1)}$

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⚝ In ∆SRP :

$\\ :\implies\:\bf\blue{tan\:x = \dfrac{10\sqrt{3}}{m + 20}} \:\dashrightarrow\:\pink{Eq^{n}\:(2)}$

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⚝ Multiplying Eqⁿ (1) and Eqⁿ (2) :

$\\ :\implies\:\sf tan\:x\:.\:cot\:x = \dfrac{10\sqrt{3}}{m + 20}\:\times\:\dfrac{10\sqrt{3}}{m}$

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• We know that, tan θ . cot θ = 1.

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Therefore,

$\\ :\implies\:\sf \dfrac{10\sqrt{3}}{m + 20}\:\times\:\dfrac{10\sqrt{3}}{m} = 1$

$\\ :\implies\:\sf \dfrac{10\:\times\:10\:{(\sqrt{3})}^{2}}{m(m + 20)} = 1$

$\\ :\implies\:\sf \dfrac{100\:\times\:3}{(m\:\times\:m) + (m\:\times\:20)} = 1$

$\\ :\implies\:\sf \dfrac{300}{m^2 + 20m} = 1$

$\\ :\implies\:\sf 300 = m^2 + 20m$

$\\ :\implies\:\sf m^2 + 20m - 300 = 0$

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Splitting the middle term :

$\\ :\implies\:\sf m^2 + (30 - 10)m - 300 = 0$

$\\ :\implies\:\sf m^2 + 30m - 10m - 300 = 0$

$\\ :\implies\:\sf m(m + 30) - 10(m + 30) = 0$

$\\ :\implies\:\sf (m + 30)\: (m - 10) = 0$

$\\ :\implies\:\sf m + 30 = 0\:\:or\:\: m - 10 = 0$

$\\ :\implies\:\sf m = 0 - 30\:\:or\:\: m = 0 + 10$

$\\ :\implies\:\underline{\boxed{\bf{\purple{m = -30\:\:or\:\: m = 10}}}}\:\bigstar$

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• m is the distance. So, it can't be negative!

• Therefore, distance b/w the building and point Q is 10 m.

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Hence,

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• Distance b/w building and point P = m + 20 = 10 + 20 = 30.

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• Therefore, distance of P from the building is 30 m.

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Note :

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• Diagram is in the attachment! $\red{\clubsuit}$

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