Question : From a point on a bridge across the river, the angles of depression of the banks on opposite sides of the river are $30^{\circ}$ and $45^{\circ}$, respectively. If the bridge is at a height of 2.5 m from the banks, then the width of the river is: Take ($\sqrt{3}$ = 1.732).
Option 1: 5.83 metres
Option 2: 6.83 metres
Option 3: 5.76 metres
Option 4: 6.87 metres
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Correct Answer: 6.83 metres
Solution :
Given,
The height of the bridge PD = 2.5 metres.
Taking one side of the river
In $\triangle$ PAD,
tan 30° = $\frac{\text{PD}}{\text{AD}}$
$\frac{1}{\sqrt{3}}$ = $\frac{2.5}{\text{AD}}$
⇒ AD = 2.5$\sqrt{3}$
Taking another side of the river
In $\Delta$ PBD,
tan 45° = $\frac{\text{PD}}{\text{BD}}$
1 = $\frac{2.5}{\text{BD}}$
⇒ BD = 2.5
$\therefore$ The width of the river is AB = AD + BD = (2.5$\sqrt{3}$ + 2.5) m.
Taking $\sqrt{3}$ = 1.732
⇒ (2.5 × 1.732) + 2.5 = 4.33 + 2.5 = 6.83 metres
Hence, the correct answer is 6.83 metres.
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