Question

from the top of tower 100m high a man observed 2 cards on opposite side of a tower with angle of depression 30 degree and 45 degree find the distance between two cars
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Answer 1

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Step-by-step explanatioYYTDSJYJn:

Answer 2

$\Large{\underline{\underline{\bf{QuEsTiOn:-}}}}$

From the top of a tower, 100 m high, a man observes two cars on the opposite sides of the tower and in same straight line with its base, with angles of depression 30 degree and 45 degree. Find the distance between the cars.

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$Let \: tower \: \\ AB = 100 \: m$

$Car \: 1 \: is \: at \: point \: c$

$Car \: 2 \: is \: at \: point \: d$

$∠ \: ACB \: = 30 {}^{o}$

$and \:  ∠ADB \: = 45 {}^{o}$

$\huge{ \red{ \bold{ \underline{ \underline{Find:-}}}}}$

The distance between the cars?

$\huge\underline\mathbb{\red S\pink {0} \purple {L} \blue {UT} \orange {1}\green {ON :}}$

$IN \: △ \: ABC \:$

$∠ABC \: = 90 {}^{o}$

$= > \:tan \: 30 = \frac{AB}{BC}$

$= > \frac{1}{ \sqrt{3} } = \frac{100} {bc} \: \: (tan \: 30 = \frac{1}{ \sqrt{3} } )$

$= > BC \: = 100 \sqrt{3 } \: m$

$IN \: △ \: ABD \:$

$∠ABD \: = 90 \: {}^{o}$

$= > tan \: 45 \: = \frac{AB}{BD}$

$= > tan \: 45 \: = \frac{1}{BD} (tan \: 45 = 1)$

$= > BD \: = 100 \: m$

$CD \: = BC \: + \: BD$

$= > CD \: = 100 \sqrt{2} + 100$

$= > 100( \sqrt{3} + 1)$

$= > 100 \: (1.732 \: + \: 1)$

$= > \: 100 \: (2.732)$

$= > \: 273.2 \: m$

$\huge\orange{\mid{\fbox{\tt{HENCE,}}\mid}}$

$Distance \: between \: \: cars \: \: is \: 273.2 \: metres.</p><p></p><p> </p><p></p><p>$