Question

The height of the tower is 100m. When the angle of elevation of the sun changes from 30 to 45 the shadow of the tower becomes x meters less. The value of x

Answer 1

$\tt\huge\underline{\underline{GIVEN}}$

• Height of tower is 100m
• Angle changed from 30° to 45° when sun's position changed.
• Shadow became x m less when sun's position changed.

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$\tt\huge\underline{\underline{TO\: FIND}}$

• Value of x

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$\tt\huge\underline{\underline{SOLUTION}}$

Let the length of shadow be S when angle is 30°

When angle is 45° then side will be S-xm.

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In ∆ACD

$\sf Tan \: 30°= \dfrac{100 \: m}{s}$

$\large\sf \implies\dfrac{1}{ \sqrt{3} } = \dfrac{100 \: m}{s}$

Now cross multiply

$\sf\large\implies S=100√3m$

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$\sf When\: angle\: is \:45°\: then\: height \:will\: be:$

$\sf\large\implies S-x m$

$\sf\large\implies 100√3-x m$

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In ∆BCD

$\sf\implies Tan \: 45° = \dfrac{100 \: m}{100 \sqrt{3} - x }$

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$\sf\large\implies 1 = \dfrac{100 \: m}{100 \sqrt{3} - x }$

Cross multiply

$\sf\large\implies100√3-x=100m$

$\sf\large\implies100√3-100=x$

$\sf\large\implies100(√3-1)=x$

$\sf\large\implies100(1.732-1)=x$

$\sf\large\implies100(0.732)=x$

$\sf\large\implies 732m=x$

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$\sf \huge\underline \blue{Additional \: information}$

TRIGONOMETRY RATIOS

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

NOTE-Kindly visit web to see diagram.

Answer 2

Answer:

$value \: of \: x \: is \: 732m$