solve for the remaining angles and side of the triangle described below. round to the nearest thousandth:

a = 100°, b = 4, c = 5

Question

solve for the remaining angles and side of the triangle described below. round to the nearest thousandth:

a = 100°, b = 4, c = 5

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 $b = 4$, $c = 5$, and $A=100^{\circ}$ into the formula:
$$
\begin{align*}
a^{2}&=4^{2}+5^{2}-2\times4\times5\times\cos(100^{\circ})\
a^{2}&=16 + 25-40\times(-0.173648)\
a^{2}&=41+6.94592\
a^{2}&=47.94592\
a&=\sqrt{47.94592}\approx6.924
\end{align*}
$$
## 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 $a\approx6.924$, $b = 4$, and $A = 100^{\circ}$ into the formula:
$$
\begin{align*}
\sin B&=\frac{b\sin A}{a}\
\sin B&=\frac{4\times\sin(100^{\circ})}{6.924}\
\sin B&=\frac{4\times0.984808}{6.924}\
\sin B&=\frac{3.939232}{6.924}\approx0.569\
B&=\sin^{-1}(0.569)\approx34.7^{\circ}
\end{align*}
$$
## Step3: Find angle $C$
Since the sum of angles in a triangle is $180^{\circ}$, $C=180^{\circ}-A - B$.
Substitute $A = 100^{\circ}$ and $B\approx34.7^{\circ}$ into the formula:
$$
\begin{align*}
C&=180^{\circ}-100^{\circ}-34.7^{\circ}\
C&=45.3^{\circ}
\end{align*}
$$

# Answer:
$a\approx6.924$, $B\approx34.7^{\circ}$, $C = 45.3^{\circ}$

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QUESTION IMAGE

Question

Explanation:

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

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

Step3: Find angle \(C\)

Answer: