In the given figure h - m-A- 30 -3- 15 -3C- 15D- 30

The correct option is B 15√(3) In Δ ADC/tan 60=h/x √(3)=h/x x=h/√(3) … (i) In Δ ABC/tan 30=h/30+x 1/√(3)=h/30+h/√(3) 1/√(3)=h√(3)/30√(3)+h 30√(3)+h=3h ...

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Standard IX

Mathematics

Angle of Elevation

In the given ...

Question

In the given figure h = ___ m.

$30 \sqrt{3}$

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$30$

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$15 \sqrt{3}$

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$15$

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Solution

The correct option is C
$15 \sqrt{3}$

$\frac{I n Δ A D C}{t a n 60 = \frac{h}{x}}$

$\sqrt{3} = \frac{h}{x}$

$x = \frac{h}{\sqrt{3}} \dots (i)$

$\frac{I n Δ A B C}{t a n 30 = \frac{h}{30 + x}}$

$\frac{1}{\sqrt{3}} = \frac{h}{30 + \frac{h}{\sqrt{3}}}$

$\frac{1}{\sqrt{3}} = \frac{h \sqrt{3}}{30 \sqrt{3} + h}$

$30 \sqrt{3} + h = 3 h$

$2 h = 30 \sqrt{3}$

$h = 15 \sqrt{3} m$

Hence (C)

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In the given figure h = ___ m.

The correct option is C 15√3 In Δ ADCtan 60=hx √3=hx x=h√3 … (i) In Δ ABCtan 30=h30+x 1√3=h30+h√3 1√3=h√330√3+h 30√3+h=3h 2h=30√3 h=15√3 m Hence (C)