Question

In △ABC, m∠A=19°, a=13, and b=14. Find c to the nearest tenth.

Answer 1

We know that
Applying the law of sines

step 1
Find the value of angle B
a=13
b=14
A=19°
so

 $\frac{a}{sin A} = \frac{b}{sin B} \\ a*sin B=b*sin A \\ sin B= \frac{b}{a}*sin A \\ sin B= \frac{14}{13} *sin 19 \\ sin B=0.3506 \\ B=arc sin(0.3506) \\ B=20.5$°

step 2
with angle A and angle B find the angle C
A+B+C=180-----> solve for C
C=180-(A+B)------> C=180-(19+20.5)-----> C=140.5°

step 3
$\frac{a}{sin A} = \frac{c}{sin C} \\ a*sin C=c*sin A \\ c=a* \frac{sin C}{sin A} \\ c=13* \frac{sin 140.5}{sin 19} \\ c=25.4$

the answer is
the value of c is 25.4