Question

Find Sin (BAC+30) 80POINTS

Answer 1

sin<BAC

• P/H
• 16/20

cos<BAC

• B/H
• 12/20

NOw

• sin(<BAC+30)
• sin<BAC×cos30+cos<BACsin30
• 16/20(√3/2)+12/20(1/2)
• 2/5√3+3/10

Answer 2

Answer:

$\dfrac{2}{5}\sqrt{3}+\dfrac{3}{10}$

Step-by-step explanation:

Trigonometric Identities

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

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

• $\theta$ 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:

• $\cos (\angle BAC)=\dfrac{12}{20}$

• $\sin (\angle BAC)=\dfrac{16}{20}$

Angles

$\sin (30^{\circ})=\dfrac{1}{2}$

$\cos (30^{\circ})=\dfrac{\sqrt{3}}{2}$

Therefore, using the trig identities and ratios:

$\begin{aligned}\sin (\angle BAC + 30^{\circ}) & = \sin (\angle BAC) \cos (30^{\circ})+\cos (\angle BAC) \sin (30^{\circ})\\\\& = \dfrac{16}{20} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{12}{20} \cdot \dfrac{1}{2}\\\\& = \dfrac{16}{40}\sqrt{3}+\dfrac{12}{40}\\\\& = \dfrac{2}{5}\sqrt{3}+\dfrac{3}{10}\end{aligned}$