Question

law of cosines: $a^2 = b^2 + c^2 - 2bccos(a)$
which equation correctly uses the law of cosines to solve for the missing side length of $\triangle pqr$?
$6^2 = p^2 + 8^2 - 2(p)(8)\cos(39^circ)$
$p^2 = 8^2 + 8^2 - 2(6)(8)\cos(39^circ)$
$8^2 = 6^2 + p^2 - 2(6)(p)\cos(39^circ)$
$p^2 = 6^2 + 6^2 - 2(6)(6)\cos(39^circ)$

Explanation:

Step1: Identify the sides and angle in the formula

In the law of cosines \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\), we want to find \(p\). So \(a = p\), \(b = 6\), \(c = 8\) and \(A=39^{\circ}\)

Step2: Substitute into the formula

Substitute \(a = p\), \(b = 6\), \(c = 8\) and \(A = 39^{\circ}\) into \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\)
We get \(p^{2}=6^{2}+8^{2}-2(6)(8)\cos(39^{\circ})\)

Answer:

\(p^{2}=6^{2}+8^{2}-2(6)(8)\cos(39^{\circ})\) (the second option)