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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Question : From the top of a 20 metres high building, the angle of elevation from the top of a tower is 60° and the angle of depression of its foot is at 45°, then the height of the tower is: $(\sqrt{3} = 1.732)$
Question : A man standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°. When he retreats 36 m from the bank, he finds that the angle is 30°. The breadth of the river is:
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