Respuesta :

Given:-

[tex]\tan A=\frac{21}{20},\cos B=\frac{15}{17}[/tex]

To find:-

[tex]\sin (A+B)[/tex]

The formula is,

[tex]\sin (A+B)=\sin A\cos B+\cos A\sin B[/tex]

Now we substitute the values. we get,

[tex]\sin (A+B)=21\times\frac{15}{17}+20\sqrt[]{(1-\frac{15^2}{17^{^2^{}}})}[/tex]

Now we get,

[tex]\begin{gathered} \sin (A+B)=21\times\frac{15}{17}+20\sqrt[]{(1-\frac{15^2}{17^{^2^{}}})} \\ \sin (A+B)=\frac{315}{17}+\frac{20}{17}\sqrt[]{17^2-15^2} \\ \sin (A+B)=\frac{315}{17}+\frac{20}{17}\times8 \\ \sin (A+B)=\frac{315}{17}+\frac{160}{17} \\ \sin (A+B)=\frac{475}{17} \end{gathered}[/tex]

So the required value is,

[tex]\frac{475}{17}[/tex]

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