Plz tell que 10 and 11 ??

Dear Student,

$$\begin{gathered} \mathrm{Suppose} \mathrm{the} \mathrm{two} \mathrm{forces} \mathrm{act} \mathrm{at} \mathrm{an} \mathrm{angle} \mathrm{\theta }, \mathrm{then} \\ \sqrt{{\mathrm{P}}^2+{\mathrm{Q}}^2}=\sqrt{(\mathrm{P}+\mathrm{Q}{)}^2+(\mathrm{P}-\mathrm{Q}{)}^2+2(\mathrm{P}+\mathrm{Q})(\mathrm{P}-\mathrm{Q}) \mathrm{cos\theta }} \\ [\mathrm{Using} \mathrm{resultant} (\mathrm{R}) \mathrm{of} \mathrm{two} \mathrm{vectors} \mathrm{A} \mathrm{and} \mathrm{B}, \mathrm{R}=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos\theta }}] \\ \mathrm{On} \mathrm{squaring}, \\ {\mathrm{P}}^2+{\mathrm{Q}}^2=(\mathrm{P}+\mathrm{Q}{)}^2+(\mathrm{P}-\mathrm{Q}{)}^2+2(\mathrm{P}+\mathrm{Q})(\mathrm{P}-\mathrm{Q}) \mathrm{cos\theta } \\ =2{\mathrm{P}}^2+2{\mathrm{Q}}^2+2({\mathrm{P}}^2-{\mathrm{Q}}^2)\mathrm{cos\theta } \\ =-2 ({\mathrm{P}}^2-{\mathrm{Q}}^2)\mathrm{cos\theta } \\ \mathrm{cos\theta }=-\frac{1}{2}\left(\frac{{\mathrm{P}}^2-{\mathrm{Q}}^2}{{\mathrm{P}}^2-{\mathrm{Q}}^2}\right)=\frac{1}{2}\left(\frac{{\mathrm{P}}^2+{\mathrm{Q}}^2}{{\mathrm{Q}}^2-{\mathrm{P}}^2}\right) \\ \mathrm{or} \mathrm{\theta }={\cos }^{-1}\left[\frac{1}{2}\right(\frac{{\mathrm{P}}^2+{\mathrm{Q}}^2}{{\mathrm{Q}}^2-{\mathrm{P}}^2}\left)\right] \\ \mathrm{Thus}, \mathrm{the} \mathrm{two} \mathrm{forces} \mathrm{act} \mathrm{at} \mathrm{an} \mathrm{angle} {\cos }^{-1}\left[\frac{1}{2}\right(\frac{{\mathrm{P}}^2+{\mathrm{Q}}^2}{{\mathrm{Q}}^2-{\mathrm{P}}^2}\left)\right]. \end{gathered}$$
Hence, the correct option is 4.

Kindly post the different question in different thread.

Regards