Question

A cricket ball is thrown over a triangle from one end of a horizontal base falls on the other end of the base after grazing the vertex. If ${\theta }_1$ and ${\theta }_2$ are the base angles of projection, then.

• A. $\tan {\theta }_1-\tan {\theta }_2=\tan \alpha$
• B. $\sin {\theta }_1+\sin {\theta }_2=\sin \alpha$
• C. $\tan {\theta }_1+\tan {\theta }_2=\tan \alpha$
• D. $\cos {\theta }_1+\cos {\theta }_2=\cos \alpha$

Answer 1

Answer: (C)

$\tan {\theta }_1+\tan {\theta }_2=\tan \alpha$

The equation of trajectory is given by
So, $y=x\tan \alpha -\frac{x^2\sin \alpha }{\left(\frac{2u^2\sin \alpha \cos \alpha }{g}\right)\cos \alpha }$
$\Rightarrow y=xtan\alpha \left(1-\frac{x}{r}\right)$ ..........(I)
Coordinates of B are $(h\cot {\theta }_1,h)$ and the range AC$=$AO$+$OC$=h\cot {\theta }_1+h\cot {\theta }_2$
Substituting the coordinate values of B in Eq. (i)
$h=h\cot {\theta }_1\tan \alpha \left(1-\frac{h\cot {\theta }_1}{h(\cot {\theta }_1+\cot {\theta }_2)}\right)$
$\Rightarrow 1=\cot {\theta }_1\cdot \tan \alpha \left[\frac{\cot {\theta }_1+\cot {\theta }_2-\cot {\theta }_1}{\cot {\theta }_1+\cot {\theta }_2}\right]$
$\Rightarrow \frac{\cot {\theta }_1+\cot {\theta }_2}{\cot {\theta }_1\cdot \cot {\theta }_2}=\tan x$
$\Rightarrow \tan {\theta }_2+\tan {\theta }_1=\tan \alpha$
$\therefore \tan {\theta }_1+\tan {\theta }_2=\tan \alpha$.