Question

solve \\( \triangle a b c \\) if \\( m \angle a = 100 ^ { \circ }, m \angle c = 23 ^ { \circ } \\), and \\( c = 250 \\) in. round each side and angle to the nearest tenth as needed. you must show all work for finding the missing sides and angles of the triangle. you must show all steps and your exact, calculator ready value before providing your rounded answer.

Explanation:

Step1: Find the measure of angle \( B \)

The sum of angles in a triangle is \( 180^{\circ} \). So, \( m\angle B=180^{\circ}-m\angle A - m\angle C \).
Substitute \( m\angle A = 100^{\circ} \) and \( m\angle C=23^{\circ} \):
\( m\angle B=180^{\circ}-100^{\circ}-23^{\circ}=57^{\circ} \)

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

The Law of Sines states that \( \frac{a}{\sin A}=\frac{c}{\sin C} \).
We know \( c = 250\), \(m\angle A=100^{\circ}\), \(m\angle C = 23^{\circ}\).
So, \( a=\frac{c\sin A}{\sin C} \).
Substitute the values: \( a=\frac{250\times\sin(100^{\circ})}{\sin(23^{\circ})} \).
Since \( \sin(100^{\circ})\approx0.9848\) and \( \sin(23^{\circ})\approx0.3907\),
\( a=\frac{250\times0.9848}{0.3907}=\frac{246.2}{0.3907}\approx630.2 \)

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

By the Law of Sines, \( \frac{b}{\sin B}=\frac{c}{\sin C} \).
We know \( c = 250\), \(m\angle B = 57^{\circ}\), \(m\angle C=23^{\circ}\).
So, \( b=\frac{c\sin B}{\sin C} \).
Since \( \sin(57^{\circ})\approx0.8387\) and \( \sin(23^{\circ})\approx0.3907\),
\( b=\frac{250\times0.8387}{0.3907}=\frac{209.675}{0.3907}\approx536.7 \)

Answer:

\( m\angle B = 57^{\circ}\), \(a\approx630.2\) in, \(b\approx536.7\) in