Question

For positive acute angles A and B, it is known that tan A = 21/20 and cos B = 15/17
Find the value of sin(A + B) in simplest form.

Answer 1

Given:-

$\tan A=\frac{21}{20},\cos B=\frac{15}{17}$

To find:-

$\sin (A+B)$

The formula is,

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

Now we substitute the values. we get,

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

Now we get,

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

So the required value is,

$\frac{475}{17}$