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From a point on a bridge $3\mathrm{m}$ above the banks, the angles of depression of the banks on opposite sides of a river are $30^{\circ}$ and $45^{\circ}$. What is the width of the river? 
A$3\sqrt{3}\mathrm{m}$
B$6\mathrm{m}$
C$3(\sqrt{3}+1)\mathrm{m}$
D$6\sqrt{3}\mathrm{m}$
Answer & Solution
Correct answer: C. $3(\sqrt{3}+1)\mathrm{m}$
Let the point vertically below the observer be $D$. Then the river width $AB=AD+DB$. Using $\tan 30^{\circ}=\frac{3}{AD}$, we get $AD=3\sqrt{3}$. Using $\tan 45^{\circ}=\frac{3}{DB}$, we get $DB=3$. Hence $AB=3\sqrt{3}+3=3(\sqrt{3}+1)\mathrm{m}$.
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