Question

the following figure shows △abc with side lengths to the nearest tenth.
find m∠c.
note that m∠c is obtuse. round to the nearest degree.
m∠c = ☐°

Explanation:

Step1: Apply Law of Sines

In $\triangle ABC$, by the Law of Sines, we have $\frac{\sin B}{AC}=\frac{\sin A}{BC}$. Let $AC = 19$, $BC = 37$, and $\angle A = 30^\circ$. So $\frac{\sin B}{19}=\frac{\sin 30^\circ}{37}$.

Step2: Solve for $\sin B$

First, calculate $\sin 30^\circ=\frac{1}{2}$. Then $\sin B=\frac{19\times\sin 30^\circ}{37}=\frac{19\times\frac{1}{2}}{37}=\frac{19}{74}\approx0.2568$.

Step3: Find $\angle B$

Take the inverse sine: $\angle B=\arcsin(0.2568)\approx14.9^\circ\approx15^\circ$.

Step4: Calculate $\angle C$

Since the sum of angles in a triangle is $180^\circ$, $\angle C = 180^\circ-\angle A-\angle B$. Substitute $\angle A = 30^\circ$ and $\angle B\approx15^\circ$: $\angle C = 180 - 30 - 15 = 135^\circ$ (we use the obtuse angle, so we consider the supplement of the acute angle from the Law of Sines, because when we use $\arcsin$, we get the acute angle, but $\angle C$ is obtuse, so we need to check the other solution for the Law of Sines. Wait, actually, when we calculated $\sin B\approx0.2568$, the other angle with the same sine is $180^\circ - 15^\circ = 165^\circ$, but then $\angle A+\angle B = 30^\circ+165^\circ = 195^\circ>180^\circ$, which is impossible. Wait, no, I made a mistake. Wait, the sides: $AC = 19$, $BC = 37$, so side opposite $\angle B$ is $AC = 19$, side opposite $\angle A$ is $BC = 37$. Wait, Law of Sines: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$, where $a$ is opposite $\angle A$, $b$ opposite $\angle B$, $c$ opposite $\angle C$. So let's correct: Let $a = BC = 37$ (opposite $\angle A$), $b = AC = 19$ (opposite $\angle B$), $c = AB$ (opposite $\angle C$). So $\frac{a}{\sin A}=\frac{b}{\sin B}\implies\frac{37}{\sin 30^\circ}=\frac{19}{\sin B}\implies\sin B=\frac{19\times\sin 30^\circ}{37}=\frac{19\times0.5}{37}=\frac{9.5}{37}\approx0.2568$. So $\angle B\approx14.9^\circ$ (acute). Then $\angle C = 180 - 30 - 14.9\approx135.1^\circ\approx135^\circ$. Wait, but why is $\angle C$ obtuse? Because the side opposite $\angle C$ (which is $AB$) should be the longest side? Wait, $BC = 37$, $AC = 19$, so $AB$: using Law of Cosines? Wait, no, let's check again. Wait, maybe I mixed up the sides. Wait, the triangle: $\angle A = 30^\circ$, side $BC = 37$ (opposite $\angle A$), side $AC = 19$ (opposite $\angle B$), side $AB$ opposite $\angle C$. So by Law of Sines, $\frac{BC}{\sin A}=\frac{AC}{\sin B}\implies\sin B=\frac{AC\times\sin A}{BC}=\frac{19\times\sin 30^\circ}{37}=\frac{19\times0.5}{37}=\frac{9.5}{37}\approx0.2568$, so $\angle B\approx15^\circ$. Then $\angle C = 180 - 30 - 15 = 135^\circ$, which is obtuse, as given. So that's correct.

Answer:

$135$