help trig

In triangle ABC, find sin A if b = 2*sqrt(10), c = 11, and a = 9.

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$a^2=b^2+c^2-2bc\ cosA\\ {\color{blue}cosA}=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{4\cdot 10+11^2-9^2}{2\cdot 11\cdot 2\cdot \sqrt{10}}=\dfrac{80}{44\cdot \sqrt 10}=\color{blue}\dfrac{20}{11\cdot \sqrt{10} }$

$sinA=\sqrt{1-cos^2A}=\sqrt{1-\dfrac{400}{121\cdot 10}}=\sqrt{\dfrac{81}{121}}\\ \color{blue}sinA= \dfrac{9}{11}=0.\overline{18}$

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