Question

using the law of sines to solve the all possible triangles if \\(\angle a = 118^\circ, a = 29, b = 18\\). if no answer exists, enter dne for all answers. \\(\angle b\\) is \\(\square\\) degrees \\(\angle c\\) is \\(\square\\) degrees \\(c = \square\\) assume \\(\angle a\\) is opposite side \\(a\\), \\(\angle b\\) is opposite side \\(b\\), and \\(\angle c\\) is opposite side \\(c\\). question help: \\(\boxed{\text{video}}\\)

Explanation:

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

The Law of Sines states that $\frac{\sin A}{a}=\frac{\sin B}{b}$.
Substituting the given values: $\frac{\sin 118^{\circ}}{29}=\frac{\sin B}{18}$.
Solving for $\sin B$: $\sin B=\frac{18\sin 118^{\circ}}{29}$.
Using a calculator, $\sin 118^{\circ}\approx0.88295$, so $\sin B=\frac{18\times0.88295}{29}\approx0.545$.
Then $B = \sin^{-1}(0.545)\approx33^{\circ}$.

Step2: Find $\angle C$

Since the sum of angles in a triangle is $180^{\circ}$, $\angle C=180^{\circ}-\angle A - \angle B$.
Substituting the values: $\angle C = 180^{\circ}-118^{\circ}-33^{\circ}=29^{\circ}$.

Step3: Find side $c$

Again using the Law of Sines: $\frac{\sin A}{a}=\frac{\sin C}{c}$.
Substituting the values: $\frac{\sin 118^{\circ}}{29}=\frac{\sin 29^{\circ}}{c}$.
Solving for $c$: $c=\frac{29\sin 29^{\circ}}{\sin 118^{\circ}}$.
Using a calculator, $\sin 29^{\circ}\approx0.4848$, so $c=\frac{29\times0.4848}{0.88295}\approx16$.

Answer:

$\angle B$ is $33$ degrees.
$\angle C$ is $29$ degrees.
$c = 16$.