Question

using the law of sines to solve the all possible triangles if $\angle a = 106^\circ, a = 30, b = 15$. if no answer exists, enter dne for all answers. $\angle b$ 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}

Explanation:

Step1: Apply the Law of Sines to find \(\angle B\)

The Law of Sines states that \(\frac{\sin A}{a}=\frac{\sin B}{b}\).
Substitute \(A = 106^{\circ}\), \(a = 30\), and \(b = 15\) into the formula:
\(\sin B=\frac{b\sin A}{a}=\frac{15\sin106^{\circ}}{30}\)
Since \(\sin106^{\circ}\approx0.9613\), then \(\sin B=\frac{15\times0.9613}{30}= 0.48065\)
\(B=\sin^{-1}(0.48065)\approx28.7^{\circ}\)

Step2: Find \(\angle C\)

Use the fact that the sum of angles in a triangle is \(180^{\circ}\), so \(C = 180^{\circ}-A - B\)
Substitute \(A = 106^{\circ}\) and \(B\approx28.7^{\circ}\)
\(C=180^{\circ}-106^{\circ}-28.7^{\circ}=45.3^{\circ}\)

Step3: Find side \(c\)

Again, use the Law of Sines \(\frac{\sin A}{a}=\frac{\sin C}{c}\)
\(c=\frac{a\sin C}{\sin A}\)
Substitute \(a = 30\), \(A = 106^{\circ}\), \(C = 45.3^{\circ}\)
Since \(\sin106^{\circ}\approx0.9613\) and \(\sin45.3^{\circ}\approx0.7107\)
\(c=\frac{30\times0.7107}{0.9613}\approx22.2\)

Answer:

\(\angle B\) is \(28.7\) degrees, \(\angle C\) is \(45.3\) degrees, \(c = 22.2\)