Question

What are the àngles of a triangle which has vertex coordinates (8, 3), (2, 9), and (1, 4)?

Answer 1

Labeling `A(8,3), B(2,9), C(1,4)` with the Law of Cosines, I get `A approx 36.87^circ`, `B approx 56.31 ^ circ`, `C approx 86.82^circ`.

Explanation:

The Law of Cosine tells us the cosine of the angle given three sides.

The Law of Cosines:

`c^2 = a^2 + b^2 - 2 a b cos C`

Solving for `cos C`:

`cos C = frac{a^2 + b^2 - c^2 }{2ab}`

Let's label the vertices `A(8,3), B(2,9), C(1,4)`

`c^2 = AB^2= 6^2+6^2=72`
`b^2 = AC^2=7^2+1^2=50`
`a^2=BC^2=1^2+5^2=26`

`cos C = {26 + 50 - 72}/{2 sqrt{50 (26) }} = \sqrt{13}/65`

`C approx 86.82^circ`

`cos B = {26 + 72 - 50}/{2 sqrt{72(26)} } = {2 sqrt{13}}/13`

`B approx 56.31 ^ circ`

`cos A = {50 + 72 -26}/{2 sqrt{72(50)}} = 4/5`

Nice rational cosine.

`A approx 36.87^circ`