(viii) (−52,5π12)(-\dfrac{5}{2},\dfrac{5\pi}{12})(−25,125π) Here r=−52<0r=-\dfrac{5}{2}<0r=−25<0 and θ=5π12\theta=\dfrac{5\pi}{12}θ=125π. Since r<0r<0r<0, use angle: 5π12+π=17π12\frac{5\pi}{12}+\pi=\frac{17\pi}{12}125π+π=1217π Use: cos5π12=6−24,sin5π12=6+24\cos\frac{5\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4},\quad \sin\frac{5\pi}{12}=\frac{\sqrt{6}+\sqrt{2}}{4}cos125π=46−2,sin125π=46+2 x=−52cos5π12=−5(6−2)8y=−52sin5π12=−5(6+2)8\begin{aligned} x&=-\frac{5}{2}\cos\frac{5\pi}{12}=-\frac{5(\sqrt{6}-\sqrt{2})}{8}\\ y&=-\frac{5}{2}\sin\frac{5\pi}{12}=-\frac{5(\sqrt{6}+\sqrt{2})}{8} \end{aligned}xy=−25cos125π=−85(6−2)=−25sin125π=−85(6+2)