From a point on a bridge across a river- the angles of depression of the banks on opposite sides of the river are 30- and 45- respectively- If the width of the river is 12 m - find the height of the bridge-

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Formula for Sum of n Terms of an AP

From a point ...

Question

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^{\circ}$ and $45^{\circ}$ , respectively. If the width of the river is 12 m , find the height of the bridge.

6( $\sqrt{3}$ - 1) m

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6 m

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6 $\sqrt{3}$ m

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3 m

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Solution

The correct option is A
6( $\sqrt{3}$ - 1) m

The situation can be represented by the figure below

$∠ P A D = ∠ A D C a n d ∠ Q A B = ∠ A B C (a l t e r n a t e a n g l e s)$
$t a n ∠ A D C = \frac{A C}{D C} = t a n 45^{\circ}$
$\Rightarrow B D = 12 c m L e t t h e l e n g t h o f B C b e x, t h e n C D = 12 - x$

$\Rightarrow t a n 45^{\circ} = \frac{A C}{12 - x}$
$\Rightarrow A C = t a n 45^{\circ} (12 - x)$
$A l s o t a n ∠ A B C = \frac{A C}{B C}$
$\Rightarrow t a n ∠ A B C = \frac{A C}{B C}$
$\Rightarrow A C = x t a n 30^{\circ}$
$\Rightarrow A C = \frac{x}{\sqrt{3}}$

$\Rightarrow A C = C D S i n c e t a n 45 = 1$
$∴ 12 - x = \frac{x}{\sqrt{3}}$
$\Rightarrow x = 6 \sqrt{3} (\sqrt{3} - 1)$
$∴ A C = x t a n 30^{\circ} = \frac{6 \sqrt{3} (\sqrt{3} - 1)}{\sqrt{3}}$
$∴ A C = 6 (\sqrt{3} - 1)$
$∴ h e i g h t o f b r i d g e i s 6 (\sqrt{3} - 1)$

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From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30∘ and 45∘, respectively. If the width of the river is 12 m , find the height of the bridge.

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^{\circ}$ and $45^{\circ}$ , respectively. If the width of the river is 12 m , find the height of the bridge.