Recall the compound angle identity:
[tex]\sin(a+b)=\sin a\cos b+\cos a\sin b[/tex]
[tex]90^\circ<a<270^\circ[/tex], so we expect [tex]\cos a<0[/tex], and [tex]0^\circ<b<180^\circ[/tex], so we expect [tex]\sin b>0[/tex]. Then by the Pythagorean identity,
[tex]\cos^2x+\sin^2x=1\implies\begin{cases}\cos a=-\sqrt{1-\sin^2a}=-\dfrac45\\\\\sin b=\sqrt{1-\cos^2b}=\dfrac{15}{17}\end{cases}[/tex]
Then
[tex]\sin(a+b)=-\dfrac{36}{85}[/tex]
