Question

Triangle ABC is formed by the points A(3, 4), B(2, 6), and C(5, 8). Find angle A to the nearest tenth of a degree?

Triangle ABC is formed by the points A(3, 4), B(2, 6), and C(5, 8). Find angle A to the nearest tenth of a degree.

Answer 1

The measure of angle `A~~53.1^@`

Explanation:

I will solve this using the Law of Cosines which states that for a triangle, with side lengths, `a`, `b`, and `c`, and angle `A` opposite the side with length `a`

`a^2=b^2+c^2-2*b*c*cos(A)`

Solving this for `cosA` we have

`A=cos^-1((b^2+c^2-a^2)/(2bc))`

Let's first calculate `a`, `b`, and `c` using the distance formula.

`a=sqrt((8-6)^2+(5-2)^2)=sqrt(13)`

`b=sqrt((8-4)^2+(5-3)^2)=2sqrt(5)`

`c=sqrt((6-4)^2+(3-2)^2)=sqrt(5)`

Plugging these values into our formula for A gives

`A=cos^-1[((2sqrt(5))^2+(sqrt(5))^2-(sqrt(13))^2)/(2*2sqrt(5)sqrt(5))]`

`=cos^-1((20+5-13)/20)=cos^-1(3/5)~~53.1^@`