Question

1. c = 18 cm, m∠a = 137°, b = 12 cm m∠b = 17°, m∠c = 26°, a = 28 cm

Explanation:

Step1: Use the Law of Cosines to verify side - length

The Law of Cosines is \(a^{2}=b^{2}+c^{2}-2bc\cos A\).
Substitute \(b = 12\), \(c = 18\), and \(A=137^{\circ}\) (where \(\cos137^{\circ}\approx - 0.7314\)) into the formula.
\[

$$\begin{align*} a^{2}&=12^{2}+18^{2}-2\times12\times18\times(-0.7314)\\ &=144 + 324+2\times12\times18\times0.7314\\ &=144+324+(432\times0.7314)\\ &=144 + 324+316.9648\\ &=784.9648 \end{align*}$$

\]
\(a=\sqrt{784.9648}\approx28\) cm.

Step2: Use the Law of Sines to verify angles

The Law of Sines is \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\).
First, check \(\frac{a}{\sin A}=\frac{b}{\sin B}\).
We know \(a = 28\), \(A = 137^{\circ}\), \(b = 12\).
\(\sin B=\frac{b\sin A}{a}=\frac{12\times\sin137^{\circ}}{28}\)
Since \(\sin137^{\circ}\approx0.682\), then \(\sin B=\frac{12\times0.682}{28}=\frac{8.184}{28}\approx0.2923\)
\(B=\arcsin(0.2923)\approx17^{\circ}\)
Then, since the sum of angles in a triangle is \(180^{\circ}\), \(C=180^{\circ}-A - B=180^{\circ}-137^{\circ}-17^{\circ}=26^{\circ}\)

Answer:

The given values of angles and side - lengths are consistent with the Laws of Sines and Cosines. So the values \(m\angle B = 17^{\circ}\), \(m\angle C=26^{\circ}\), \(a = 28\) cm are correct for the triangle with \(c = 18\) cm, \(m\angle A=137^{\circ}\), \(b = 12\) cm.