Question

using the law of sines to solve the triangle if $\angle a = 38^\circ, \angle c = 75^\circ, b = 11$ :
$\angle b$ is \boxed{} degrees;
$a = \boxed{}$ ;
$c = \boxed{}$ ;
you may round to two decimal places.
assume $\angle a$ is opposite side $a$, $\angle b$ is opposite side $b$, and $\angle c$ is opposite side $c$.
question help: \video

Explanation:

Step1: Find angle \( B \)

The sum of angles in a triangle is \( 180^{\circ} \). So \( \angle B=180^{\circ}-\angle A - \angle C \).
Substitute \( \angle A = 38^{\circ} \) and \( \angle C=75^{\circ} \):
\( \angle B=180^{\circ}-38^{\circ}-75^{\circ}=67^{\circ} \)

Step2: Use the Law of Sines to find \( a \)

The Law of Sines is \( \frac{a}{\sin A}=\frac{b}{\sin B} \).
We know \( b = 11 \), \( \angle A=38^{\circ} \), \( \angle B = 67^{\circ} \).
So \( a=\frac{b\sin A}{\sin B} \).
\( \sin A=\sin(38^{\circ})\approx0.6157 \), \( \sin B=\sin(67^{\circ})\approx0.9205 \)
\( a=\frac{11\times0.6157}{0.9205}=\frac{6.7727}{0.9205}\approx7.36 \)

Step3: Use the Law of Sines to find \( c \)

By the Law of Sines \( \frac{c}{\sin C}=\frac{b}{\sin B} \).
\( \sin C=\sin(75^{\circ})\approx0.9659 \), \( \sin B=\sin(67^{\circ})\approx0.9205 \), \( b = 11 \)
\( c=\frac{b\sin C}{\sin B}=\frac{11\times0.9659}{0.9205}=\frac{10.6249}{0.9205}\approx11.54 \)

Answer:

\(\angle B = 67\) degrees; \(a\approx7.36\); \(c\approx11.54\)