Find the interior angle of the triangle with vertices A=(3,2) B=(4,5) C=(-1,-1)

Question

And please can you show me the solution. Thank you in advance. :)

Answer 1

There are three interior angles, all of which can be found.

First, the length of the sides. CA=5, because we can use Pythagoras and the coordinates and of points A and C:

AC^2=(2-(-1))^2+(3-(-1))^2=9+16, so AC=sqrt(25)=5.

Similarly, BC=sqrt(61) and AB=sqrt(10), using the coords of B and C, and A and B.

Using the cosine rule, we can find angle A:

a^2=b^2+c^2-2bccosA:

61=25+10-2*5sqrt(10)cosA; cosA=-26/(10sqrt(10))=-0.8222, A=145.30 degrees.

We can now use the sine rule to find angle B:

sinB/b=sinA/a:

sinB=5*0.5692/sqrt(61)=0.3644, B=21.37 degrees. Angle C=180-(A+B)=13.33 degrees.