Answer:
[tex]\dfrac{2}{5}\sqrt{3}+\dfrac{3}{10}[/tex]
Step-by-step explanation:
Trigonometric Identities
[tex]\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B[/tex]
Trigonometric ratios
[tex]\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}[/tex]
where:
- [tex]\theta[/tex] is the angle
- O is the side opposite the angle
- A is the side adjacent the angle
- H is the hypotenuse (the side opposite the right angle)
Using the trig ratio formulas for cosine and sine:
- [tex]\cos (\angle BAC)=\dfrac{12}{20}[/tex]
- [tex]\sin (\angle BAC)=\dfrac{16}{20}[/tex]
Angles
[tex]\sin (30^{\circ})=\dfrac{1}{2}[/tex]
[tex]\cos (30^{\circ})=\dfrac{\sqrt{3}}{2}[/tex]
Therefore, using the trig identities and ratios:
[tex]\begin{aligned}\sin (\angle BAC + 30^{\circ}) & = \sin (\angle BAC) \cos (30^{\circ})+\cos (\angle BAC) \sin (30^{\circ})\\\\& = \dfrac{16}{20} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{12}{20} \cdot \dfrac{1}{2}\\\\& = \dfrac{16}{40}\sqrt{3}+\dfrac{12}{40}\\\\& = \dfrac{2}{5}\sqrt{3}+\dfrac{3}{10}\end{aligned}[/tex]
