Find the length XZ in the triangle

Question 1 Report

√6m

\sqrt{5 m}

\sqrt{3 m}

Answer Details

xz2 = xy2 + yx2 - 2(xy) (yz) cos 120 $^{\circ}$

= 22 + 12 - 2(2) (1) cos 1202

= 4 + 1 - 4x - cos 60 = 5 - 4x - $\frac{1}{2}$

5 + 2 = 7

xz = $\sqrt{7}$ m