Question

If $3p$ and $4p$ are resultant of a force $5p$ , then the angle between $3p$ and $5p$ is

• A. ${\sin }^{-1}\left(\frac{3}{5}\right)$
• B. ${\sin }^{-1}\left(\frac{4}{5}\right)$
• C. $90^{\circ }$
• D. None of these

Answer 1

Answer: (B)

$Q=\frac{R\sin \alpha }{\sin \left(\alpha +\beta \right)}$
Also , ${\left(5P\right)}^2={\left(4P\right)}^2+{\left(3P\right)}^2+2\left(4P\right)\left(3P\right)\cos \left(\alpha +\beta \right)$
$\Rightarrow 25P^2=16p^2+9P^2+24P^2\cos \left(\alpha +\beta \right)$
$\Rightarrow 24P^2\cos \left(\alpha +\beta \right)=0$
$\Rightarrow \cos \left(\alpha +\beta \right)=0=\cos 90^{\circ }$
$\Rightarrow \alpha +\beta =90^{\circ }$

Now, $4P=\frac{5P\sin \alpha }{\sin 90^{\circ }}$
$\Rightarrow \sin \alpha =\frac{4}{5}$
$\Rightarrow \alpha ={\sin }^{-1}\left(\frac{4}{5}\right)$