Question

solve the triangle.
a=8.199 in c=6.602 in b=76.58°
what is the length of side b?
□ in
(round to the nearest thousandth as needed.)
what is the measure of angle a?
□°
(round to the nearest hundredth as needed.)
what is the measure of angle c?
□°
(round to the nearest hundredth as needed.)

Explanation:

Step1: Find side b using the Law of Cosines

The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and the angle \(B\) opposite side \(b\), \(b^{2}=a^{2}+c^{2}-2ac\cos(B)\).
Given \(a = 8.199\) in, \(c=6.602\) in, \(B = 76.58^{\circ}\).
First, calculate \(a^{2}=(8.199)^{2}\approx67.2236\), \(c^{2}=(6.602)^{2}\approx43.5864\), and \(2ac\cos(B)=2\times8.199\times6.602\times\cos(76.58^{\circ})\).
\(\cos(76.58^{\circ})\approx0.2325\), so \(2\times8.199\times6.602\times0.2325\approx2\times8.199\times6.602\times0.2325\approx2\times8.199\times1.535\approx2\times12.586\approx25.172\).
Then \(b^{2}=67.2236 + 43.5864- 25.172=85.638\). So \(b=\sqrt{85.638}\approx9.254\) in.

Step2: Find angle A using the Law of Sines

The Law of Sines states that \(\frac{\sin(A)}{a}=\frac{\sin(B)}{b}\).
We know \(a = 8.199\), \(b\approx9.254\), \(B = 76.58^{\circ}\).
So \(\sin(A)=\frac{a\sin(B)}{b}=\frac{8.199\times\sin(76.58^{\circ})}{9.254}\).
\(\sin(76.58^{\circ})\approx0.9703\), so \(\sin(A)=\frac{8.199\times0.9703}{9.254}\approx\frac{7.956}{9.254}\approx0.859\).
Then \(A=\arcsin(0.859)\approx59.13^{\circ}\).

Step3: Find angle C

Since the sum of angles in a triangle is \(180^{\circ}\), \(C=180^{\circ}-A - B\).
We know \(A\approx59.13^{\circ}\), \(B = 76.58^{\circ}\), so \(C=180-(59.13 + 76.58)=180 - 135.71 = 44.29^{\circ}\).

Answer:

Length of side \(b\): \(\boxed{9.254}\) in
Measure of angle \(A\): \(\boxed{59.13}\)°
Measure of angle \(C\): \(\boxed{44.29}\)°