quiz 8: solve right, scalene, and obtuse triangles
6 points possible answered: 4/6
question 5
solve for the remaining sides and angles if possible (if not, answer \dne\ in all answer boxes). as in the text, (a, a), (b, b), and (c, c) are angle - side opposite pairs. round to one decimal places, if necessary.
a = 110°, b = 30, c = 28
a =
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b =
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c =
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a =
b =
c =
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Accepted Answer

# Explanation:
## Step1: Use the Law of Cosines to find side $a$
The Law of Cosines formula is $a^{2}=b^{2}+c^{2}-2bc\cos A$.
Substitute $b = 30$, $c = 28$, and $A=110^{\circ}$ into the formula:
$$
\begin{align*}
a^{2}&=30^{2}+28^{2}-2\times30\times28\times\cos(110^{\circ})\
a^{2}&=900 + 784-1680\times(-0.3420)\
a^{2}&=1684+ 574.56\
a^{2}&=2258.56\
a&=\sqrt{2258.56}\approx47.5
\end{align*}
$$
## Step2: Use the Law of Sines to find angle $B$
The Law of Sines formula is $\frac{\sin B}{b}=\frac{\sin A}{a}$.
Substitute $b = 30$, $a\approx47.5$, and $A = 110^{\circ}$ into the formula:
$$
\begin{align*}
\sin B&=\frac{b\sin A}{a}\
\sin B&=\frac{30\times\sin(110^{\circ})}{47.5}\
\sin B&=\frac{30\times0.9397}{47.5}\
\sin B&=\frac{28.191}{47.5}\approx0.5935\
B&=\sin^{-1}(0.5935)\approx36.4^{\circ}
\end{align*}
$$
## Step3: Use the angle - sum property of a triangle to find angle $C$
Since $A + B+C=180^{\circ}$, then $C=180^{\circ}-A - B$.
Substitute $A = 110^{\circ}$ and $B\approx36.4^{\circ}$ into the formula:
$$
\begin{align*}
C&=180^{\circ}-110^{\circ}-36.4^{\circ}\
C&=33.6^{\circ}
\end{align*}
$$

# Answer:
$A = 110^{\circ}$, $B\approx36.4^{\circ}$, $C\approx33.6^{\circ}$, $a\approx47.5$, $b = 30$, $c = 28$

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Question

Explanation:

Step1: Use the Law of Cosines to find side \(a\)

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

Step3: Use the angle - sum property of a triangle to find angle \(C\)

Answer: