Question

From a balloon vertically above a straight road, the angle of depression of two cars at an instant is found to be ${45}^0$ and ${60}^0$. If the cars are $100$ m apart, then the height of the balloon is $\frac{m}{2}$. Find $m$.

Answer 1

REF Image according to the question.

$b$ is the height of the balloon. $a$ is the distance of 1st car having an angle of depression ${60}^o$ from the balloon.

Now,

$tan{45}^0=\left(\frac{b}{a+100}\right)\to \left(1\right)tan{60}^0=\left(\frac{b}{a}\right)\to \left(2\right)\therefore \left(\frac{tan{45}^0}{tan{60}^0}\right)=\left(\frac{a}{a+100}\right)=\left(\frac{1}{\sqrt{3}}\right)=\left(\frac{a}{a+100}\right)a+100=\sqrt{3}a\therefore a=\left(\frac{100}{\sqrt{3}-1}\right)=50(\sqrt{3}+1)\therefore b=a+100=50(\sqrt{3}+1)+100=236.6=\left(\frac{473}{2}\right)\left(approx\right)\therefore m=473$ $b=a+100$ from eq(1)