The range of $f\left(x\right)=\sqrt{9{\sin }^{2}x-30\sin x+25+}10$ is $\left[a,b\right]$ then $a+b=$

$\sqrt{9{\sin }^{2}x-30\sin x+25}+10$

$=\sqrt{{(3\sin x-5)}^2}+10=|3\sin x-5|+10$

$-1\le \sin x\le 1\Rightarrow -3\le 3\sin x\le 3$

$\Rightarrow -8\le 3\sin x-5\le -2$

$\Rightarrow 2\le |3\sin x-5|\le 8$

$\Rightarrow 12\le |3\sin x-5|+10\le 18$

$\therefore a+b=12+18=30$