Question

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

Explanation:

Step1: Match law of cosines terms

The law of cosines is $a^2 = b^2 + c^2 - 2bc\cos(A)$, where $a$ is the side opposite angle $A$, and $b,c$ are the sides adjacent to angle $A$.

Step2: Identify triangle components

In $\triangle PQR$, angle $P = 39^\circ$, adjacent sides are $PR=6$ and $PQ=8$, opposite side is $QR=p$.

Step3: Substitute into formula

Substitute $a=p$, $b=6$, $c=8$, $A=39^\circ$ into the law of cosines.
$p^2 = 6^2 + 8^2 - 2(6)(8)\cos(39^\circ)$

Answer:

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