Question

suppose a triangle has two sides of length 42 and 35, and that the angle between these two sides is 120°. which equation should you solve to find the length of the third side of the triangle?
a. $c^{2}=42^{2}+35^{2}-2(42)(35)cos120^{circ}$
b. $c^{2}=42^{2}+35^{2}-2(42)(35)sin120^{circ}$
c. $\frac{sin42}{120}=\frac{sin35}{b}$
d. $c = 42 + 35 - 2(42)(35)cos120^{circ}$

Explanation:

Step1: Recall the law of cosines

The law of cosines for a triangle with sides \(a\), \(b\), \(c\) and the included - angle \(C\) between sides \(a\) and \(b\) is \(c^{2}=a^{2}+b^{2}-2ab\cos C\).

Step2: Identify the values of \(a\), \(b\), and \(C\)

Let \(a = 42\), \(b = 35\), and \(C=120^{\circ}\). The side we want to find is \(c\).

Step3: Substitute the values into the law - of - cosines formula

Substituting the values into the formula \(c^{2}=a^{2}+b^{2}-2ab\cos C\), we get \(c^{2}=42^{2}+35^{2}-2(42)(35)\cos120^{\circ}\).

Answer:

A. \(c^{2}=42^{2}+35^{2}-2(42)(35)\cos120^{\circ}\)