Question

solve the triangle.
a = 1090 in, b = 855 in, c = 773 in
what is the degree measure of angle a?
(simplify your answer. type an integer or decimal rounded to the nearest tenth as needed.)
what is the degree measure of angle b?
(simplify your answer. type an integer or decimal rounded to the nearest tenth as needed.)
what is the degree measure of angle c?
(simplify your answer. type an integer or decimal rounded to the nearest tenth as needed.)

Explanation:

Step1: Use the Law of Cosines for angle A

The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) opposite angles \(A\), \(B\), \(C\) respectively, \(\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}\).
Given \(a = 1090\), \(b = 855\), \(c = 773\).
First, calculate \(b^{2}+c^{2}-a^{2}\):
\(b^{2}=855^{2}=731025\), \(c^{2}=773^{2}=597529\), \(a^{2}=1090^{2}=1188100\)
\(b^{2}+c^{2}-a^{2}=731025 + 597529-1188100=1328554 - 1188100 = 140454\)
Then, calculate \(2bc\): \(2\times855\times773 = 2\times661965=1323930\)
So, \(\cos A=\frac{140454}{1323930}\approx0.1061\)
Then, \(A=\arccos(0.1061)\approx83.9^{\circ}\)

Step2: Use the Law of Cosines for angle B

\(\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}\)
Calculate \(a^{2}+c^{2}-b^{2}\): \(1188100+597529 - 731025=1785629 - 731025 = 1054604\)
Calculate \(2ac\): \(2\times1090\times773=2\times842570 = 1685140\)
\(\cos B=\frac{1054604}{1685140}\approx0.626\)
Then, \(B=\arccos(0.626)\approx51.2^{\circ}\)

Step3: Use the fact that the sum of angles in a triangle is \(180^{\circ}\) for angle C

Since \(A + B + C=180^{\circ}\), then \(C=180^{\circ}-A - B\)
\(C=180-(83.9 + 51.2)=180 - 135.1 = 44.9^{\circ}\)

Answer:

Angle A: \(83.9^{\circ}\)
Angle B: \(51.2^{\circ}\)
Angle C: \(44.9^{\circ}\)