Question

using the law of sines to solve the all possible triangles if $\angle a = 119^\circ, a = 29, b = 16$. 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} submit question

Explanation:

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

The Law of Sines states that \(\frac{\sin A}{a}=\frac{\sin B}{b}\).
Substituting the given values \(A = 119^{\circ}\), \(a = 29\), and \(b = 16\) into the formula:
\(\sin B=\frac{b\sin A}{a}=\frac{16\sin119^{\circ}}{29}\)
\(\sin119^{\circ}=\sin(180 - 61)^{\circ}=\sin61^{\circ}\approx0.8746\)
\(\sin B=\frac{16\times0.8746}{29}\approx\frac{13.9936}{29}\approx0.4825\)
\(B=\sin^{- 1}(0.4825)\approx28.8^{\circ}\)

Step2: Find \(\angle C\)

Since the sum of angles in a triangle is \(180^{\circ}\), \(C = 180^{\circ}-A - B\)
\(C=180^{\circ}-119^{\circ}-28.8^{\circ}=32.2^{\circ}\)

Step3: Find side \(c\)

Using the Law of Sines \(\frac{\sin A}{a}=\frac{\sin C}{c}\)
\(c=\frac{a\sin C}{\sin A}\)
\(\sin C=\sin32.2^{\circ}\approx0.532\), \(\sin A=\sin119^{\circ}\approx0.8746\)
\(c=\frac{29\times0.532}{0.8746}\approx\frac{15.428}{0.8746}\approx17.64\)

Answer:

\(\angle B\) is \(28.8\) degrees, \(\angle C\) is \(32.2\) degrees, \(c = 17.64\)