Question

in $\triangle def$, $de = 11$, $ef = 9$, and angle $e = 140^{\circ}$.
which equation correctly uses the law of cosines to
solve for the third side?
$e^{2} = 11^{2} + 9^{2} - 2(11)(9)\cos(140^{\circ})$
$11^{2} = e^{2} + 9^{2} - 2e(9)\cos(140^{\circ})$
$9^{2} = e^{2} + 11^{2} - 2e(11)\cos(140^{\circ})$
$e = 11 + 9 - 2(11)(9)\cos(140^{\circ})$

Explanation:

Step1: State the Law of Cosines formula

The Law of Cosines formula is \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\), where \(a\) is the side opposite angle \(A\), and \(b\) and \(c\) are the other two sides.

Step2: Identify the sides and angle in \(\triangle DEF\)

In \(\triangle DEF\), let \(a = e\), \(b = 11\) (side \(DE\)), \(c = 9\) (side \(EF\)), and \(A=140^{\circ}\) (angle \(E\)).

Step3: Substitute into the Law of Cosines formula

Substituting the values into \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\), we get \(e^{2}=11^{2}+9^{2}-2(11)(9)\cos(140^{\circ})\).

Answer:

\(e^{2}=11^{2}+9^{2}-2(11)(9)\cos(140^{\circ})\) (the first option).