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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[tex]\large \bf \clubs \: Question :- [/tex]
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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[tex]\large \bf \clubs \: Given :- [/tex]
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[tex]\large \bf \clubs \: To \: Find :- [/tex]
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[tex]\large \bf \clubs \: Answer :- [/tex]
Distance of point P from building is 30m.
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[tex]\large \bf \clubs \:Solution:- [/tex]
As the angles are complementary for their sum must be 90°.
According to the Attached Figure:
[tex]\large \bf In \: \triangle \: SRQ[/tex]
[tex]\sf \tan(90 - x) = \dfrac{10 \sqrt{3} }{y} \\ \\ \bf :\longmapsto cot \: x = \frac{10 \sqrt{3} }{y} \: \: \: \:. \: . \: . \: (i) [/tex]
Now ,
[tex]\bf \large In \: \triangle \: SRP [/tex]
[tex]\bf tan \: x = \dfrac{10 \sqrt{3} }{y + 20} \: \: \: \: . \: .\: . \: (ii)[/tex]
Mulyiplying (i) and (ii) We get,
[tex] \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 } }}}[/tex]
As y is distance So it can't be Negative
Hence,
[tex]\Large\purple{ :\longmapsto \underline {\boxed{{\bf y = 10m} }}}[/tex]
So,
Distance of point P from building
= 10 + 20 meters
That is
[tex]\pink{ \LARGE \mathfrak{Distance=30m}}[/tex]
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Given :
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To Find :
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Solution :
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⚝ Things to know before solving :
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⚝ In ∆SRQ :
[tex]\\ :\implies\:\sf tan(90^{\circ} - x) = \dfrac{SR}{RQ} [/tex]
[tex]\\ :\implies\:\bf\pink{cot\:x = \dfrac{10\sqrt{3}}{m}} \:\dashrightarrow\:\blue{Eq^{n}\:(1)}[/tex]
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⚝ In ∆SRP :
[tex]\\ :\implies\:\bf\blue{tan\:x = \dfrac{10\sqrt{3}}{m + 20}} \:\dashrightarrow\:\pink{Eq^{n}\:(2)}[/tex]
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⚝ Multiplying Eqⁿ (1) and Eqⁿ (2) :
[tex]\\ :\implies\:\sf tan\:x\:.\:cot\:x = \dfrac{10\sqrt{3}}{m + 20}\:\times\:\dfrac{10\sqrt{3}}{m} [/tex]
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Therefore,
[tex]\\ :\implies\:\sf \dfrac{10\sqrt{3}}{m + 20}\:\times\:\dfrac{10\sqrt{3}}{m} = 1 [/tex]
[tex]\\ :\implies\:\sf \dfrac{10\:\times\:10\:{(\sqrt{3})}^{2}}{m(m + 20)} = 1[/tex]
[tex]\\ :\implies\:\sf \dfrac{100\:\times\:3}{(m\:\times\:m) + (m\:\times\:20)} = 1[/tex]
[tex]\\ :\implies\:\sf \dfrac{300}{m^2 + 20m} = 1 [/tex]
[tex]\\ :\implies\:\sf 300 = m^2 + 20m[/tex]
[tex]\\ :\implies\:\sf m^2 + 20m - 300 = 0[/tex]
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Splitting the middle term :
[tex]\\ :\implies\:\sf m^2 + (30 - 10)m - 300 = 0[/tex]
[tex]\\ :\implies\:\sf m^2 + 30m - 10m - 300 = 0[/tex]
[tex]\\ :\implies\:\sf m(m + 30) - 10(m + 30) = 0[/tex]
[tex]\\ :\implies\:\sf (m + 30)\: (m - 10) = 0[/tex]
[tex]\\ :\implies\:\sf m + 30 = 0\:\:or\:\: m - 10 = 0[/tex]
[tex]\\ :\implies\:\sf m = 0 - 30\:\:or\:\: m = 0 + 10[/tex]
[tex]\\ :\implies\:\underline{\boxed{\bf{\purple{m = -30\:\:or\:\: m = 10}}}}\:\bigstar[/tex]
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Hence,
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Note :
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