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Let $\mathrm{A}$ and $\mathrm{B}$ be obtuse angles such that $\sin \mathrm{A}=\frac{4}{5}$ and
$\cos B=-\frac{12}{13}$. What is the value of $\sin (A+B) ?$
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$\cos B=-\frac{12}{13}$. What is the value of $\sin (A+B) ?$
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The correct answer is:
$-\frac{63}{65}$
$\sin \mathrm{A}=\frac{4}{5}$ and $\cos \mathrm{B}=-\frac{12}{13}$
It is given that $\mathrm{A}$ and $\mathrm{B}$ are obtuse angle
$\Rightarrow \cos \mathrm{A}-\pm \sqrt{1-\sin ^{2} \mathrm{~A}}-\pm \sqrt{1-\frac{16}{25}}--\frac{3}{5}$
Negative sign is taken for cos A since A being obtuse lies in second quadrant. $\sin B=\pm \sqrt{1-\cos ^{2} B}=\pm \sqrt{1-\left(\frac{-12}{13}\right)^{2}}$
$=\sqrt{\frac{169-144}{169}}=\frac{5}{13}$
Positive sign is taken since, $\sin B$ is positive in second quadrant.
$\Rightarrow \cos \mathrm{A}=\frac{-3}{5}$ and $\sin \mathrm{B}=\frac{3}{13}$
$\therefore \sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$
$=\frac{4}{5} \times\left(\frac{-12}{13}\right)+\frac{-3}{5} \times\left(\frac{5}{13}\right)=-\frac{48}{65}-\frac{15}{65}$
$=\frac{-48-15}{65}=\frac{-63}{65}$
It is given that $\mathrm{A}$ and $\mathrm{B}$ are obtuse angle
$\Rightarrow \cos \mathrm{A}-\pm \sqrt{1-\sin ^{2} \mathrm{~A}}-\pm \sqrt{1-\frac{16}{25}}--\frac{3}{5}$
Negative sign is taken for cos A since A being obtuse lies in second quadrant. $\sin B=\pm \sqrt{1-\cos ^{2} B}=\pm \sqrt{1-\left(\frac{-12}{13}\right)^{2}}$
$=\sqrt{\frac{169-144}{169}}=\frac{5}{13}$
Positive sign is taken since, $\sin B$ is positive in second quadrant.
$\Rightarrow \cos \mathrm{A}=\frac{-3}{5}$ and $\sin \mathrm{B}=\frac{3}{13}$
$\therefore \sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$
$=\frac{4}{5} \times\left(\frac{-12}{13}\right)+\frac{-3}{5} \times\left(\frac{5}{13}\right)=-\frac{48}{65}-\frac{15}{65}$
$=\frac{-48-15}{65}=\frac{-63}{65}$
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