Question

part 1 of 3
solve the triangle.
a=3, b=7, c=10°

c ≈
(do not round until the final answer. then round to the nearest tenth as needed.)

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines for a triangle with sides \(a\), \(b\), \(c\) and the angle \(C\) opposite side \(c\) is \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). Here, \(a = 3\), \(b = 7\), and \(C=10^{\circ}\). Substitute these values into the formula:
\(c^{2}=3^{2}+7^{2}-2\times3\times7\times\cos(10^{\circ})\)
Calculate each term: \(3^{2}=9\), \(7^{2}=49\), \(2\times3\times7 = 42\), and \(\cos(10^{\circ})\approx0.9848\).
So \(c^{2}=9 + 49-42\times0.9848\)

Step2: Simplify the expression

First, calculate \(9 + 49=58\). Then, \(42\times0.9848\approx41.3616\).
So \(c^{2}=58 - 41.3616=16.6384\)

Step3: Find the value of \(c\)

Take the square root of \(c^{2}\): \(c=\sqrt{16.6384}\approx4.1\) (rounded to the nearest tenth)

Answer:

\(4.1\)