A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60- when he retires 40 feet from the bank he finds the angle to be 30- Find the height of the tree and the breadth of the river-

Let h be the #160; height of tree, then #160; h 40 =sin 60^ o ∴ #160;h=20√( 3 ) ft. h BQ =tan 60^ o =√( 3 ),BQ= h √( (3) )=20 ft. = ...

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Standard X

Mathematics

Calculating Heights and Distances

A person stan...

Question

A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is $60^{\circ}$ , when he retires $40$ feet from the bank he finds the angle to be $30^{\circ}$ . Find the height of the tree and the breadth of the river.

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Solution

Let h be the height of tree, then
$\frac{h}{40} = sin 60^{o}$
$∴ h = 20 \sqrt{3} f t .$
$\frac{h}{B Q} = tan 60^{o} = \sqrt{3}, B Q = \frac{h}{\sqrt{(3)}} = 20 f t .$
$=$ breadth of the river.

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A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60∘, when he retires 40 feet from the bank he finds the angle to be 30∘. Find the height of the tree and the breadth of the river.

Let h be the height of tree, then h40=sin60o∴h=20√3 ft.hBQ=tan60o=√3,BQ=h√(3)=20 ft.=breadth of the river.

A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is $60^{\circ}$ , when he retires $40$ feet from the bank he finds the angle to be $30^{\circ}$ . Find the height of the tree and the breadth of the river.

Let h be the height of tree, then
$\frac{h}{40} = sin 60^{o}$
$∴ h = 20 \sqrt{3} f t .$
$\frac{h}{B Q} = tan 60^{o} = \sqrt{3}, B Q = \frac{h}{\sqrt{(3)}} = 20 f t .$
$=$ breadth of the river.