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 bridge is at a height of 3 m from the banks, find the width of the river.
Step-by-Step Solution
Key Concept: Use the definition of angle of depression and the tangent ratio in right‑angled triangles. For an angle of depression $\theta$, $\tan\theta = \dfrac{\text{height}}{\text{horizontal distance}}$. Hence horizontal distances to each bank are obtained as $\dfrac{h}{\tan\theta}$ and summed to get the total width.
1. Draw a horizontal line through the point $P$ on the bridge (height $h=3\,\text{m}$).\
2. Drop perpendiculars from $P$ to the two banks meeting the river banks at $A$ (near bank) and $B$ (far bank).\
3. $\angle$ of depression to the near bank $A$ is $45^{\circ}$, to the far bank $B$ is $30^{\circ}$.\
4. In right‑angled triangle $\triangle P A Q$ (where $Q$ is the foot of the perpendicular on the near bank),\
$$\tan 45^{\circ}=\frac{\text{height}}{PA}=\frac{3}{PA}\;\Rightarrow\;PA=\frac{3}{\tan45^{\circ}}=\frac{3}{1}=3\,\text{m}.$$\
5. In right‑angled triangle $\triangle P B R$ (where $R$ is the foot of the perpendicular on the far bank),\
$$\tan 30^{\circ}=\frac{\text{height}}{PB}=\frac{3}{PB}\;\Rightarrow\;PB=\frac{3}{\tan30^{\circ}}=\frac{3}{\frac{1}{\sqrt{3}}}=3\sqrt{3}\,\text{m}.$$\
6. The total width of the river $= PA + PB = 3 + 3\sqrt{3}=3(1+\sqrt{3})\,\text{m}.$
Correct Answer: $3(1+\sqrt{3})\ \text{metres}\;\approx\;8.20\ \text{m}$