Question

If $\tan 5A=\tan \theta$ and $\tan (3A+B)=1$, and $QA=B+\theta -\pi /p,n\in Z$. Also it is given that the value of $\sqrt{P^3+R\cdot Q^2}=10$, then find the value of $R$

Answer 1

Answer: 6

$\tan 5A=\tan \theta$
$\Rightarrow 5A=n\pi +\theta$
$\Rightarrow A=\frac{n\pi }{5}+\frac{\theta }{5}$
$\tan (3A+B)=1$
$\Rightarrow \tan (3A+B)=\tan \frac{\pi }{4}$
$\Rightarrow 3A+B=n\pi +\frac{\pi }{4}$
$\Rightarrow B=n\pi -\frac{3n\pi }{5}-\frac{3\theta }{5}+\frac{\pi }{4}=\frac{2n\pi }{5}-\frac{3\theta }{5}+\frac{\pi }{4}$
By the equation $QA=B+\theta -\frac{\pi }{p}(n\in Z)$
$\Rightarrow \frac{Qn\pi }{5}+\frac{Q\theta }{5}=\frac{2n\pi }{5}-\frac{3\theta }{5}+\frac{\pi }{4}+\theta -\frac{\pi }{p}$
$\Rightarrow \frac{Qn\pi }{5}+\frac{Q\theta }{5}=\frac{2n\pi }{5}+\frac{2\theta }{5}+\frac{\pi }{4}-\frac{\pi }{p}$
$\Rightarrow Q=2,p=4$,
Also $\sqrt{p^3+R\cdot Q^2}=10$
$\Rightarrow \sqrt{64+R\cdot 4}=10$
$\Rightarrow R=9$