The problem in the image asks to solve for the missing sides $a$ and $b$ of a right triangle given:

• One of the angles is $50^\circ$,
• The length of the side opposite the right angle (the hypotenuse) is $17$.

We can solve for the sides using trigonometric ratios, specifically sine and cosine functions:

1. For side $a$ (opposite the $50^\circ$ angle): $\sin(50^\circ) = \frac{a}{17}$ Solving for $a$: $a = 17 \times \sin(50^\circ)$

2. For side $b$ (adjacent to the $50^\circ$ angle): $\cos(50^\circ) = \frac{b}{17}$ Solving for $b$: $b = 17 \times \cos(50^\circ)$

Let me calculate these values for you.The calculated lengths of the triangle's sides are:

• $a \approx 13.0$ (opposite the $50^\circ$ angle),
• $b \approx 10.9$ (adjacent to the $50^\circ$ angle).

These values are rounded to the nearest tenth, as requested.

Would you like more details on the steps or have any further questions?

Here are some related questions:

1. How can the Pythagorean theorem be used to check the calculations?
2. What are the definitions of sine, cosine, and tangent in right triangles?
3. How would the values change if the angle was $40^\circ$ instead of $50^\circ$?
4. How does the triangle’s angle sum property apply to right triangles?
5. What are the trigonometric ratios for angles in a 45-45-90 triangle?

Tip: For solving right triangles, always start with the angle and hypotenuse when using trigonometric ratios.