Question

abby used the law of cosines for $\triangle kmn$ to solve for $k$.
$k^2 = 31^2 + 53^2 - 2(31)(53)\cos(37\degree)$

law of cosines: $a^2 = b^2 + c^2 - 2bc\cos(a)$

what additional information did abby know that is not shown in the diagram?
\bigcirc $m\angle k = 37\degree$ and $n = 31$
\bigcirc $m\angle k = 37\degree$ and $k = 31$
\bigcirc $m\angle n = 37\degree$ and $n = 31$
\bigcirc $m\angle n = 37\degree$ and $k = 31$

Explanation:

The law of cosines is \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\), where \(A\) is the angle opposite side \(a\), and \(b\) and \(c\) are the other two sides. In the formula \(k^{2}=31^{2}+53^{2}-2(31)(53)\cos(37^{\circ})\), comparing with the law of cosines formula, we can see that:

• The side opposite the angle of \(37^{\circ}\) is \(k\) (so \(a = k\)).
• The other two sides are \(31\) and \(53\). From the diagram, side \(KM\) is labeled \(n\), so \(n = 31\) (since one of the sides is \(31\)).
• The angle of \(37^{\circ}\) should be the angle between the sides of length \(31\) and \(53\). In triangle \(KMN\), the angle between \(KM\) (length \(n = 31\)) and \(KN\) (length \(53\)) is \(\angle K\). So \(\angle K=37^{\circ}\) and \(n = 31\).

Let's analyze each option:

• Option 1: \(m\angle K = 37^{\circ}\) and \(n = 31\) – This matches our analysis.
• Option 2: \(m\angle K = 37^{\circ}\) and \(k = 31\) – But \(k\) is the side we are solving for, and from the formula, \(k\) is opposite the \(37^{\circ}\) angle, so \(k\) can't be equal to \(31\) (since \(31\) is one of the other sides), so this is incorrect.
• Option 3: \(m\angle N = 37^{\circ}\) and \(n = 31\) – \(\angle N\) is not the angle between the sides of length \(31\) and \(53\), so this is incorrect.
• Option 4: \(m\angle N = 37^{\circ}\) and \(k = 31\) – Similar to option 2, \(k\) is not equal to \(31\) and \(\angle N\) is not the angle between the sides of length \(31\) and \(53\), so this is incorrect.

Answer:

A. \(m\angle K = 37^{\circ}\) and \(n = 31\)