P3=Q5=R4\frac{P}{3}=\frac{Q}{5}=\frac{R}{4}3P=5Q=4R হলে PR=?P^R=?PR=?
2π3\frac{2\pi}{3} 32π
π2\frac{\pi}{2} 2π
cos−1(−1114)\cos^{-1}{\left(-\frac{11}{14}\right)} cos−1(−1411)
cos−1(1114)\cos^{-1}{\left(\frac{11}{14}\right)} cos−1(1411)
cosθ=32+42−522×3×4=0∴θ=π2\cos{\theta}=\frac{3^2+4^2-5^2}{2\times3\times4}=0\therefore\theta=\frac{\pi}{2} cosθ=2×3×432+42−52=0∴θ=2π
