(C) Let $OP$ be the height of the tree and $OA$ be the breadth of the river.In $\triangle OAP$,$\tan 60^{\circ} = \frac{OP}{OA} = \sqrt{3} \implies OP

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(C) Let $OP$ be the height of the tree and $OA$ be the breadth of the river.In $\triangle OAP$,$\tan 60^{\circ} = \frac{OP}{OA} = \sqrt{3} \implies OP = OA \sqrt{3}$.When the person moves $40 \ m$ away,the new distance from the bank is $OB = OA + 40$.In $\triangle OBP$,$\tan 30^{\circ} = \frac{OP}{OB} = \frac{1}{\sqrt{3}}$.Substituting $OP = OA \sqrt{3}$ into the second equation:$\frac{OA \sqrt{3}}{OA + 40} = \frac{1}{\sqrt{3}}$$3 OA = OA + 40$$2 OA = 40$$OA = 20 \ m$.

Competitive Quantitative Aptitude Heights and Distances Heights and Distances

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$A$ person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank is $60^{\circ}$. When he moves $40 \ m$ away from the bank,he finds the angle of elevation to be $30^{\circ}$. The breadth of the river is (in $m$):

A
$40$
B
$60$
C
$20$
D
$30$

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Heights and Distances

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Heights and Distances

$A$ $10 \ m$ long ladder is placed against a wall. It is inclined at an angle of $30^{\circ}$ to the ground. The distance (in $m$) of the foot of the ladder from the wall is (Given $\sqrt{3} = 1.732$)

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From a point $20 \; m$ away from the foot of a tower,the angle of elevation of the top of the tower is $30^{\circ}$. The height of the tower is:

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From the top of a $25 \ m$ high cliff,the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find the height of the tower in $m$.

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Two towers of equal height stand on either side of a wide road which is $100 \, m$ wide. At a point on the road between the towers, the angles of elevation of the tops of the towers are $60^{\circ}$ and $30^{\circ}$. Find their heights in metres. (in $\sqrt{3}$)

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$A$ man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes $12$ minutes for the angle of depression to change from $30^{\circ}$ to $45^{\circ},$ how soon after this will the car reach the observation tower?

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$A$ person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank is $60^{\circ}$. When he moves $40 \ m$ away from the bank,he finds the angle of elevation to be $30^{\circ}$. The breadth of the river is (in $m$):

Similar Questions

$A$ $10 \ m$ long ladder is placed against a wall. It is inclined at an angle of $30^{\circ}$ to the ground. The distance (in $m$) of the foot of the ladder from the wall is (Given $\sqrt{3} = 1.732$)

From a point $20 \; m$ away from the foot of a tower,the angle of elevation of the top of the tower is $30^{\circ}$. The height of the tower is:

From the top of a $25 \ m$ high cliff,the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find the height of the tower in $m$.

Two towers of equal height stand on either side of a wide road which is $100 \, m$ wide. At a point on the road between the towers, the angles of elevation of the tops of the towers are $60^{\circ}$ and $30^{\circ}$. Find their heights in metres. (in $\sqrt{3}$)

$A$ man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes $12$ minutes for the angle of depression to change from $30^{\circ}$ to $45^{\circ},$ how soon after this will the car reach the observation tower?

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