Angle between vectors

Question

The points 𝐴(2; -1), 𝐵(3; 5) and 𝐶(-4; 2) are given.

Find (a) the sum 0,5𝐶𝐴⃗+ 0,4𝐶𝐵⃗ , (b) the lengths of the sides of triangle ABC, (c) the angles of triangle ABC, (d) area of triangle ABC.

Answer 1

vector CA is a vector from origin (-6,3)

vector CB from origin (-7,-3)

.5CA+.4CB=(-5.8,.3)

AB=√37

AC=√45

BC=√58

law of cosines

cosA=[58-37-45]/[-2√37√45]

cosB=[45-37-58/{[-2√37√58]

cosC=(37-45-58]/[-2√45√58]

area=(1/2)(AB)()AC)sinA

since cosA=12/[√37√45]

sinA=[√(37*45-12²)]/[√37√45]

so area=(1/2)(√(37*45-144)=39/2

is the answer sufficient?