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# From a balloon vertically above a straight road, the angles of depression of two cars at an instant are found to be 45° and 60°.If the cars are 100m apart, then the height of the balloon is $\dfrac{m}{2}$ . Find m.

Last updated date: 12th Aug 2024
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Hint: We can draw a diagram marking all the angles formed with respect to the balloon that is placed vertically above a straight road. Then by finding the value of tan of angles of respective triangles formed, we can calculate the height of the balloon. In this value, we can then substitute the height as $\dfrac{m}{2}$ so as to calculate the value of m.
$\tan \theta = \dfrac{P}{B}$ where, $\theta$ is the angle formed, P is the Perpendicular and B is the base for the respective angle.

Let A be the balloon vertically above a straight road and P and Q be the two cars on the road. The angles of depression are formed between the imaginary straight line from A and the lines joining A to the respective cars. The angles of same measure as their corresponding angles are formed on the road as the lines are parallel to each other. The diagram can be given as:

In $\vartriangle AQB$ , the tan of angle measuring 60° can be given as:
$\tan \theta = \dfrac{P}{B}$ here,
Angle $\left( \theta \right)$ = 60°
Perpendicular (P) = AB
Base (B) = QB
$\Rightarrow \tan {60^\circ } = \dfrac{{AB}}{{QB}} \\ \Rightarrow \sqrt 3 = \dfrac{h}{x} \\ \Rightarrow x = \dfrac{h}{{\sqrt 3 }}...(1) \\ \left( \because \tan {60^\circ } = \sqrt 3 , AB = h, QB = x \right) \;$
In $\vartriangle APB$ , the tan of angle measuring 45° can be given as:
$\tan \theta = \dfrac{P}{B}$ here,
Angle $\left( \theta \right)$ = 45°
Perpendicular (P) = AB
Base (B) = BP
$\Rightarrow \tan {45^\circ } = \dfrac{{AB}}{{BP}} \\ \Rightarrow 1 = \dfrac{h}{{100 + x}} \\ \therefore 100 + x = h...(2) \\ \left( \because \tan {60^\circ } = \sqrt 3, AB = h , BP = 100 + x \right) \;$
We can find the value of h using (1) and (2):
Substituting the value of x from (1) in (2), we get:
$\Rightarrow 100 + \dfrac{h}{{\sqrt 3 }} = h \\ \Rightarrow h - \dfrac{h}{{\sqrt 3 }} = 100 \\ \Rightarrow \dfrac{{h\sqrt 3 - h}}{{\sqrt 3 }} = 100 \\ \Rightarrow h\left( {\sqrt 3 - 1} \right) = 100\sqrt 3 \\ \therefore h = \dfrac{{100\sqrt 3 }}{{\sqrt 3 - 1}} \;$
Rationalising this value of h:
$h = \dfrac{{100\sqrt 3 }}{{\sqrt 3 - 1}} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}} \\ h = \dfrac{{100\left( {3 + \sqrt 3 } \right)}}{{{{\left( {\sqrt 3 - 1} \right)}^2}}}.....(3) \\ h = \dfrac{{100\left( {3 + \sqrt 3 } \right)}}{2} \\ h = 50\left( {3 + \sqrt 3 } \right) \\ h = 150 + 50\sqrt 3 \;$
Now, it is given that for the given conditions, the height of the balloon is $\dfrac{m}{2}$
$\Rightarrow h = \dfrac{m}{2}$
Substituting this value in (3) so as to calculate the value of m:
$\dfrac{m}{2} = 150 + 50\sqrt 3 \\ \Rightarrow m = 2\left( {150 + 50\sqrt 3 } \right) \\ \therefore m = 300 + 100\sqrt 3 \;$
Therefore, for the given conditions, the value of m is $300 + 100\sqrt 3$
So, the correct answer is “ $300 + 100\sqrt 3$ ”.

Note: For any angle of a triangle, the perpendicular is the side opposite to it, base is the line on which it is formed and the longest side is hypotenuse. There was a need to rationalize the obtained value of h because we generally try to avoid the square roots in the denominator. For rationalization, remember:
We multiply the same quantity present in the denominator but with opposite sign by both numerator and denominator. This makes the square root in the denominator as natural as it gets squared.