Given:
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.
The bridge is at a height of 3 m from the banks,
Concept Used:
0° 
30° 
45° 
60° 
90° 

sin 
0 
\(\frac{1}{2}\) 
\(\frac{1}{\sqrt2}\) 
\(\frac{\sqrt3}{2}\) 
1 
cos 
1 
\(\frac{\sqrt3}{2}\) 
\(\frac{1}{\sqrt2}\) 
\(\frac{1}{2}\) 
0 
tan 
0 
\(\frac{1}{\sqrt3}\) 
1 
√3 
∞ 
Calculation:
Let AB be the banks of the river and P be the point on the bridge.
Draw PM ⊥ AB.
In ΔAMP, tan 45° = \(\rm\frac{PM}{AM}\).
⇒ 1 = \(\rm\frac{3}{AM}\)
⇒ AM = 3 m.
In ΔBMP, tan 30° = \(\rm\frac{PM}{BM}\).
⇒ \(\rm\frac{1}{\sqrt3}=\frac{3}{BM}\)
⇒ BM = 3√3 m.
Now, AB = AM + BM = 3 + 3√3 = 3(1 + √3) m.
Hence, the width of the river is 3(√3 + 1) m.