solve for the remaining angles and side of the triangle described below. round to the nearest thousandth: b = 45°, a = 6, c = 5

Question

Accepted Answer

# Explanation:
## Step1: Find side $b$ using the Law of Cosines
The Law of Cosines formula is $b^{2}=a^{2}+c^{2}-2ac\cos B$.
Substitute $a = 6$, $c = 5$, and $B=45^{\circ}$ ($\cos45^{\circ}=\frac{\sqrt{2}}{2}\approx0.7071$) into the formula:
$$
\begin{align*}
b^{2}&=6^{2}+5^{2}-2\times6\times5\times0.7071\
b^{2}&=36 + 25-42.426\
b^{2}&=18.574\
b&=\sqrt{18.574}\approx  4.310
\end{align*}
$$
## Step2: Find angle $A$ using the Law of Sines
The Law of Sines formula is $\frac{\sin A}{a}=\frac{\sin B}{b}$.
We know $a = 6$, $b\approx4.310$, and $B = 45^{\circ}$ ($\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.7071$).
$$
\begin{align*}
\sin A&=\frac{a\sin B}{b}\
\sin A&=\frac{6\times0.7071}{4.310}\
\sin A&=\frac{4.2426}{4.310}\approx0.9844\
A&=\sin^{-1}(0.9844)\approx79.7^{\circ}
\end{align*}
$$
## Step3: Find angle $C$
Since the sum of angles in a triangle is $A + B+C=180^{\circ}$.
$$
\begin{align*}
C&=180^{\circ}-A - B\
C&=180^{\circ}-79.7^{\circ}-45^{\circ}\
C&=55.3^{\circ}
\end{align*}
$$

# Answer:
$b\approx4.310$, $A\approx79.7^{\circ}$, $C\approx55.3^{\circ}$

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QUESTION IMAGE

Question

Explanation:

Step1: Find side \(b\) using the Law of Cosines

Step2: Find angle \(A\) using the Law of Sines

Step3: Find angle \(C\)

Answer: