Question

AV10С3BFind sin(2-ZBAC).

Answer 1

Trigonometric ratios for angle BAC:

$\begin{gathered} \text{sin}\angle\text{BAC}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{3}{\sqrt[]{10}} \\ \\ \text{cos}\angle\text{BAC}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{1}{\sqrt[]{10}} \end{gathered}$

Use the trigonometric ratios above to solve sin (2BAC):

$\begin{gathered} \sin 2x=2\cdot\sin x\cdot\cos x \\ \\ \sin (2\cdot\angle\text{BAC)}=2\cdot\sin \angle BAC\cdot\cos \angle BAC \\ \\ \sin (2\cdot\angle\text{BAC)}=2\cdot\frac{3}{\sqrt[]{10}}\cdot\frac{1}{\sqrt[]{10}} \\ \\ \sin (2\cdot\angle\text{BAC)}=\frac{2\cdot3\cdot1}{(\sqrt[]{10})^2} \\ \\ \sin (2\cdot\angle\text{BAC)}=\frac{6}{10} \\ \\ \sin (2\cdot\angle\text{BAC)}=\frac{3}{5} \end{gathered}$