Question

I need help with my pre-calculus homework, please show me how to solve them step by step if possible. I have tried "one solution" but it was incorrect. Please let me know if it would be "two solutions" or "no solutions" and how.

Answer 1

We are given BC = 16, AC = 5, and the measure of angle B = 42°.

Applying the law of sines:

`\frac{16}{sin\text{ A}}=\frac{5}{sin\text{ B}}`

Where A is the angle in vertex A.

Solving for sin A:

`sin\text{ A}=\frac{16\text{ }sin\text{ 42\degree}}{5}=(16*0.669)/(5)=2.14`

We have the sine of angle A with a value that is greater than 1. The sine can only have values between -1 and 1, so this triangle has no solution.