Question

which equation correctly uses the law of cosines to solve for y? law of cosines: $a^2 = b^2 + c^2 - 2bccos(a)$ triangle xyz with xz = y, xy = 9, yz = 19, angle at y is 41°. options: $9^2 = y^2 + 19^2 - 2(y)(19)cos(41°)$; $y^2 = 9^2 + 19^2 - 2(y)(19)cos(41°)$; $9^2 = y^2 + 19^2 - 2(9)(19)cos(41°)$; $y^2 = 9^2 + 19^2 - 2(9)(19)cos(41°)$

Explanation:

Step1: Recall Law of Cosines

The law of cosines is given by \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\), where \(a\) is the side opposite angle \(A\), and \(b\) and \(c\) are the other two sides of the triangle.

Step2: Identify Sides and Angle

In triangle \(XYZ\), we want to solve for \(y\) (side \(XZ\)). The angle opposite to side \(y\) is the angle at \(Y\), which is \(41^{\circ}\). The other two sides adjacent to angle \(Y\) are \(XY = 9\) and \(YZ=19\). So, if we let \(a = y\), \(b = 9\), \(c = 19\), and \(A=41^{\circ}\), substituting into the law of cosines formula:

\(y^{2}=9^{2}+19^{2}-2(9)(19)\cos(41^{\circ})\)

Answer:

\(y^{2}=9^{2}+19^{2}-2(9)(19)\cos(41^{\circ})\) (the fourth option: \(y^{2}=9^{2}+19^{2}-2(9)(19)\cos(41^{\circ})\))