QuestionUsing the Law of Sines to solve the all possible triangles if $\angle A=110^{\circ}, a=29, b=15$. If no answer exists, enter DNE for all answers. $\angle B$ is $\square$ degrees $\angle C$ is $\square$ degrees $$c=$$ $\square$

Studdy Solution
Calculate side $c$ using the Law of Sines:
$\frac{a}{\sin A} = \frac{c}{\sin C}$
Substitute the known values:
$\frac{29}{\sin 110^\circ} = \frac{c}{\sin 41.0^\circ}$
Solve for $c$:
$c = \frac{29 \cdot \sin 41.0^\circ}{\sin 110^\circ}$
Calculate $c$:
$c \approx \frac{29 \cdot 0.6561}{0.9397} \approx 20.3$
The solution for the triangle is:
$\angle B = 29.0^\circ$
$\angle C = 41.0^\circ$
$c \approx 20.3$