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find the missing side - and angle - measures in triangle abc. round you…

Question

find the missing side - and angle - measures in triangle abc. round your answers to the nearest tenth.
the measure of angle b is approximately
the measure of angle c is approximately
the length of side ab is approximately

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. In $\triangle ABC$, we know $A = 40^{\circ}$, $b = 18$, and $c=23$. First, $\frac{BC}{\sin A}=\frac{AC}{\sin B}=\frac{AB}{\sin C}$.

Step2: Find angle $B$

$\frac{18}{\sin B}=\frac{23}{\sin40^{\circ}}$. Then $\sin B=\frac{18\times\sin40^{\circ}}{23}$.
$\sin B=\frac{18\times0.6428}{23}=\frac{11.57}{23}\approx0.503$. So $B=\sin^{- 1}(0.503)\approx30.2^{\circ}$.

Step3: Find angle $C$

Since the sum of angles in a triangle is $180^{\circ}$, $C = 180^{\circ}-A - B$. So $C=180^{\circ}-40^{\circ}-30.2^{\circ}=109.8^{\circ}$.

Step4: Find side $AB$

Using the Law of Sines again, $\frac{AB}{\sin C}=\frac{AC}{\sin B}$. So $AB=\frac{18\times\sin109.8^{\circ}}{\sin30.2^{\circ}}$.
$\sin109.8^{\circ}\approx0.94$, $\sin30.2^{\circ}\approx0.503$. Then $AB=\frac{18\times0.94}{0.503}=\frac{16.92}{0.503}\approx33.6$.

Answer:

The measure of angle $B$ is approximately $30.2^{\circ}$
The measure of angle $C$ is approximately $109.8^{\circ}$
The length of side $AB$ is approximately $33.6$