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 it retreats $20m$ from the bank, he finds the angle to be $30^\circ $ then the height of the tree and the breadth of the river is:
A. $10\sqrt 3 m,10m$
B. $10,10\sqrt 3 m$
C. $20m,30m$
D. none of these

Answer 1

Hint:
Let us assume the height of the tree be $h$ and the breadth of the river be $x$
Now draw the diagram using the above statement. Two right angle triangles will be formed and in any right angle triangle $\tan \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$
Using this in the two triangles we will get our answer.

Complete step by step solution:
Here in the question we are given that person standing on the bank of a river, observes that the angle subtended by a tree on the opposite bank is $60^\circ $ when it retreats $20m$ from the bank, he finds the angle to be $30^\circ $
Now let us assume that height of the tree be $h$ and the breadth of the river be $x$
Now let us understand by drawing the diagram

Here in the figure we can see that $BC$ is the tree of height $h$ so it means that $BC = h$ and $AB$ is the breadth of the river which means$AB = x$.
Now initially the person is at point A and he observes that the tree which is $BC$ subtends the angle of $60^\circ $ so we can say that $\angle BAC = 60^\circ $
Now we are also given in the question that when it is $20m$ away from the bank then he reaches at the point D and then he observes the angle subtended to be $30^\circ $on the opposite bank so it means that $\angle CDB = 30^\circ $
Now it is clear from the figure that there are two right angles triangles which are $\Delta ABC,\Delta DBC$
 $\tan A = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$$ = \dfrac{{BC}}{{AB}} = \dfrac{h}{x}$
And $\angle A = 60^\circ $ so $\tan 60^\circ = \dfrac{h}{x}$
So we get that $\sqrt 3 x = h - - - - - - (1)$
Now in $\Delta DBC$
$\tan D = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$$ = \dfrac{{BC}}{{BD}} = \dfrac{h}{{20 + x}}$
So we get that $\tan 30^\circ = \dfrac{h}{{20 + x}}$
$\dfrac{1}{{\sqrt 3 }} = \dfrac{h}{{20 + x}}$
$\Rightarrow 20 + x = \sqrt 3 h - - - - - - (2)$
And we know that $\sqrt 3 x = h$
So putting the value of $h$in the equation (2)
$\Rightarrow 20 + x = \sqrt 3 (\sqrt 3 x)$
$\Rightarrow 20 + x = 3x$
$\Rightarrow 2x = 20$
$\Rightarrow x = 10$
So we get that $h = \sqrt 3 x = 10\sqrt 3 m$
So we get that breadth of the river is $10m$ and the height of the tree as $10\sqrt 3 m$

So option A is correct.

Note:
For any triangle $\Delta ABC$

Given that $\angle B = 90^\circ ,\angle A = \theta $
$\sin \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}$$ = \dfrac{{BC}}{{AC}}$
And perpendicular is the length opposite to that of the angle and hypotenuse is the side opposite to $90^\circ $ and similarly we must know the formula of other trigonometric functions.