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प्रश्न
The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's elevation is 30°than when it was 45°. The height of the tower in metres is
पर्याय
\[\left( \sqrt{3} + 1 \right) x\]
\[\left( \sqrt{3} - 1 \right) x\]
\[2\sqrt{3}x\]
\[3\sqrt{2}x\]
उत्तर
Let h be the height of tower AB
Given that: angle of elevation of sun are`∠D=30°` and.`∠C=45°`
Then Distance`CD=2x` and we assume `BC=x`Here, we have to find the height of tower.
So we use trigonometric ratios.
In a triangle,`ABC`
`⇒ tan C=(AB)/(BC)`
`⇒ tan 45°=(AB)/(BC)`
`⇒ 1=h/x`
`⇒ x=h`
Again in a triangle ABD,
`⇒ tan D= (AB)/(BC+CD)`
`⇒ tan 30°=h/( x+2x)`
`⇒1/sqrt3=h/(h+2x)` `[x=h]`
`⇒ sqrt3h=h+2x`
`h(sqrt3-1)=2x`
`⇒ h=2x/(sqrt3-1)`
⇒` h=(2x)/(sqrt3-1)xx(sqrt3+1)/(sqrt3+1)`
`⇒ h=x(sqrt3+1)`
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