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 25 P^{2} = 16 p^{2} + 9P^{2} +24 P^{2} \cos\left(\alpha + \beta\right) $
$\Rightarrow 24 P^{2} \cos\left(\alpha + \beta\right) = 0$
$ \Rightarrow \cos\left(\alpha + \beta\right) = 0 = \cos90^{\circ} $
$ \Rightarrow \alpha + \beta = 90^{\circ}$

Now, $ 4P = \frac{5P \sin\alpha}{\sin90^{\circ}}$
$ \Rightarrow \sin\alpha = \frac{4}{5}$
$\Rightarrow \alpha=\sin^{-1} \left(\frac{4}{5}\right) $