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(A) Let $CD$ be the tower of height $60 \ m$ and $AB$ be the other tower of height $h$. Let the width of the river be $BD = x$.From the diagram,$CD =

Vedclass · Accepted Answer

$20 \sqrt{3}, 40$

Vedclass · Answer

(A) Let $CD$ be the tower of height $60 \ m$ and $AB$ be the other tower of height $h$. Let the width of the river be $BD = x$.From the diagram,$CD = 60 \ m$. The angle of depression of the top of tower $AB$ (point $A$) from $C$ is $30^{\circ}$,and the angle of depression of the bottom of tower $AB$ (point $B$) from $C$ is $60^{\circ}$.In $\Delta CBD$,$	an 60^{\circ} = \frac{CD}{BD} = \frac{60}{x}$.Since $	an 60^{\circ} = \sqrt{3}$,we have $\sqrt{3} = \frac{60}{x}$,which gives $x = \frac{60}{\sqrt{3}} = 20 \sqrt{3} \ m$. Thus,the width of the river is $20 \sqrt{3} \ m$.Now,draw a horizontal line from $A$ to $CD$,meeting $CD$ at point $E$. Then $AE = BD = 20 \sqrt{3} \ m$ and $ED = AB = h$.In $\Delta CAE$,$	an 30^{\circ} = \frac{CE}{AE} = \frac{CE}{20 \sqrt{3}}$.Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{CE}{20 \sqrt{3}}$,which gives $CE = 20 \ m$.Therefore,the height of the other tower $AB = ED = CD - CE = 60 - 20 = 40 \ m$.

Two towers are situated on opposite banks of a river. One tower is $60 \ m$ high. From the top of this tower,the angles of depression of the top and the bottom of the other tower are $30^{\circ}$ and $60^{\circ}$ respectively. Find the width of the river and the height of the other tower (in $m$).

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