Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 325: 80

$A = 9.5$

$A = (2,5)$ $B = (-1,3)$ $C = (4,0)$ We can find the area of the triangle: $A = \frac{1}{2}\vert (x_1y_2-y_1x_2+x_2y_3-y_2x_3+x_3y_1-y_3x_1) \vert$ $A = \frac{1}{2}\vert [(2)(3)-(5)(-1)+(-1)(0)-(3)(4)+(4)(5)-(0)(2)] \vert$ $A = \frac{1}{2}\vert [(6)-(-5)+(0)-(12)+(20)-(0)] \vert$ $A = \frac{1}{2}\vert 19 \vert$ $A = 9.5$