1. **State the problem:** We are given values of $x$ and $y$ that satisfy the equation $$y = a \sin x^\circ + b$$ and need to find $y$ when $x = 45^\circ$. 2. **Use the given data to find $a$ and $b$:** From the table: - When $x=0^\circ$, $y=5$. - When $x=90^\circ$, $y=7$. Substitute $x=0^\circ$ into the equation: $$y = a \sin 0^\circ + b = a \times 0 + b = b$$ So, $$b = 5$$. Substitute $x=90^\circ$: $$7 = a \sin 90^\circ + 5 = a \times 1 + 5$$ Solve for $a$: $$a = 7 - 5 = 2$$. 3. **Verify with another point:** At $x=30^\circ$, $y=6$. Check: $$6 \stackrel{?}{=} 2 \sin 30^\circ + 5 = 2 \times \frac{1}{2} + 5 = 1 + 5 = 6$$ This confirms our values. 4. **Find $y$ when $x=45^\circ$:** $$y = 2 \sin 45^\circ + 5 = 2 \times \frac{\sqrt{2}}{2} + 5 = \cancel{2} \times \frac{\sqrt{2}}{\cancel{2}} + 5 = \sqrt{2} + 5$$ **Final answer:** $$y = 5 + \sqrt{2}$$