Question

using the law of sines to solve for all possible triangles if $\angle a = 113^{\circ}$, $a = 26$, $b = 18$. assume $\angle a$ is opposite side $a$, $\angle b$ is opposite side $b$, and $\angle c$ is opposite side $c$. if no triangles exist, enter dne for all answers. if multiple triangles exist, input all angles and sides as a list. round your solutions to the nearest tenth place. a solution write - up is required for this problem. answers: $\angle b = \square$ $\angle c = \square$ $c = \square$

Explanation:

Step1: Apply Law of Sines for ∠B

The Law of Sines states $\frac{a}{\sin A}=\frac{b}{\sin B}$. We know $A = 113^\circ$, $a = 26$, $b = 18$. So, $\sin B=\frac{b\sin A}{a}=\frac{18\sin113^\circ}{26}$. Calculate $\sin113^\circ\approx\sin(90^\circ + 23^\circ)=\cos23^\circ\approx0.9205$. Then $\sin B\approx\frac{18\times0.9205}{26}\approx\frac{16.569}{26}\approx0.6373$. Now, find $B$: $B\approx\arcsin(0.6373)\approx39.6^\circ$ (since $A$ is obtuse, $B$ must be acute, so only one solution for $B$).

Step2: Calculate ∠C

Since the sum of angles in a triangle is $180^\circ$, $C = 180^\circ - A - B\approx180^\circ - 113^\circ - 39.6^\circ = 27.4^\circ$.

Step3: Apply Law of Sines for side c

Using $\frac{a}{\sin A}=\frac{c}{\sin C}$, so $c=\frac{a\sin C}{\sin A}$. We know $a = 26$, $\sin C\approx\sin27.4^\circ\approx0.4609$, $\sin A\approx0.9205$. Then $c\approx\frac{26\times0.4609}{0.9205}\approx\frac{11.9834}{0.9205}\approx13.0$.

Answer:

$\angle B\approx39.6^\circ$, $\angle C\approx27.4^\circ$, $c\approx13.0$