find the interior angles of the triangle whose vertices are A (4, 3) , B(-2,2) and C(2,-8).
Answers
Answered by
6
Step-by-step explanation:
When plotting the points and applying distance formula to find lengths, we find that:
AB = sqrt(37)
BC = sqrt(116)
AC = sqrt(125)
Via law of cosines:
AB^2 = BC^2 + AC^2 - 2(BC)(AC)cos(t) where t is the angle between sides AC and BC.
37 = 116 + 125 - 2*sqrt(116*125)*cos(t)
t = 32.1 degrees (Angle C)
Now use law of sines:
sqrt(37)/sin(32.1) = sqrt(125)/sin(B)
B = 77.615 degrees
Finally angle A = 180 - 77.615 - 32.1 = 70.285 degrees.
Similar questions