\(R=\sqrt{A^2+B^2+2 A B \cos \theta}\)
\(A=8, B=15, R=17\)
\(17^2=8^2+15^2+2 \times 8 \times 15 \times \cos \theta\)
\(289=64+225+240 \cos \theta\)
\(\Rightarrow 289=289+24 \cos \theta\)
\(24 \cos \theta=0\)
\(\cos \theta=0 \Rightarrow \theta=90^{\circ}\)