Question

using the law of sines to solve for all possible triangles if $\angle b = 50^\circ, a = 108, b = 42$. if no answer exists, enter dne for all answers. $\angle a$ is \boxed{} degrees $\angle c$ is \boxed{} degrees $c = \boxed{}$ assume $\angle a$ is opposite side $a$, $\angle b$ is opposite side $b$, and $\angle c$ is opposite side $c$. question help: \boxed{video} submit question

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that \(\frac{\sin A}{a}=\frac{\sin B}{b}\).
We know that \(B = 50^{\circ}\), \(a = 108\), and \(b = 42\).
Substituting the values into the formula: \(\sin A=\frac{a\sin B}{b}=\frac{108\times\sin50^{\circ}}{42}\).
Calculate \(\sin50^{\circ}\approx0.7660\), then \(\sin A=\frac{108\times0.7660}{42}\approx\frac{82.728}{42}\approx1.9697\).
Since the range of the sine function is \([- 1,1]\), and \(1.9697>1\), no such triangle exists.

Answer:

\(\angle A=\text{DNE}\)
\(\angle C=\text{DNE}\)
\(c = \text{DNE}\)