Question

determine the remaining sides and angles of the triangle abc.
a = 109.63°, c = 30.53°, c = 250
b = □°
a ≈ □
(do not round until the final answer. then round to the nearest tenth as needed.)
b ≈ □
(do not round until the final answer. then round to the nearest tenth as needed.)

Explanation:

Step1: Find angle B

The sum of angles in a triangle is \(180^\circ\). So, \(B = 180^\circ - A - C\).
Substitute \(A = 109.63^\circ\) and \(C = 30.53^\circ\):
\(B = 180 - 109.63 - 30.53 = 39.84^\circ\approx39.8^\circ\) (rounded to nearest tenth)

Step2: Find side a using the Law of Sines

Law of Sines: \(\frac{a}{\sin A}=\frac{c}{\sin C}\)
So, \(a = \frac{c \cdot \sin A}{\sin C}\)
Substitute \(c = 250\), \(A = 109.63^\circ\), \(C = 30.53^\circ\):
\(\sin A=\sin(109.63^\circ)\approx\sin(180 - 70.37)^\circ=\sin(70.37^\circ)\approx0.942\)
\(\sin C=\sin(30.53^\circ)\approx0.508\)
\(a=\frac{250\times0.942}{0.508}=\frac{235.5}{0.508}\approx463.6\)

Step3: Find side b using the Law of Sines

Law of Sines: \(\frac{b}{\sin B}=\frac{c}{\sin C}\)
So, \(b = \frac{c \cdot \sin B}{\sin C}\)
\(\sin B=\sin(39.84^\circ)\approx0.641\)
\(b=\frac{250\times0.641}{0.508}=\frac{160.25}{0.508}\approx315.4\)

Answer:

\(B = \boxed{39.8}^\circ\)
\(a \approx \boxed{463.6}\)
\(b \approx \boxed{315.4}\)