In this problem, we are given a right triangle with one angle $50^\circ$, and the side opposite the right angle ($AB$) labeled as $18$. We are asked to solve for side $b$ (the base) and side $c$ (the hypotenuse) of the triangle, and the other angle, $B = 40^\circ$, has been provided in the box. We'll use trigonometric functions to solve this.

Step 1: Use Trigonometric Ratios

• We know angle $A = 50^\circ$ and side $AB = 18$.

1. Solving for side $c$ (the hypotenuse):

We can use the sine function because: $\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{18}{c}$ So, $c = \frac{18}{\sin(50^\circ)}$ $c \approx \frac{18}{0.7660} \approx 23.5$ Thus, $c \approx 23.5$.

2. Solving for side $b$ (the base):

We can use the tangent function because: $\tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{18}{b}$ So, $b = \frac{18}{\tan(50^\circ)}$ $b \approx \frac{18}{1.1918} \approx 15.1$ Thus, $b \approx 15.1$.

Final Results:

• $b \approx 15.1$
• $c \approx 23.5$

Would you like more details or have any questions on this?

Here are 5 related questions:

1. How do you use the cosine function to verify the base $b$?
2. What happens if you round the values to different decimal places?
3. How can you find angle $B$ if it wasn't given?
4. How does the Pythagorean theorem apply to this triangle?
5. Can you solve this triangle if a different side is given?

Tip: Always double-check if your calculator is in degrees mode when working with angles in degrees!