# Thread: Determine if this is a right triangle.

1. ## Determine if this is a right triangle.

Given A(5,2), B(-3,6), and C(1,-3), determine if triangle ABC is a right triangle.

So first I took a look at the angle between A and B:

cos theta = [5,2][-3,6] / magnitude of [5,2] x magnitude of [-3,6]

= -15 + 12 / square root of 25+4 x square root of 9+36

= -3 / square root of 1305

Then I took a look at the angle between B and C:

cos theta = [-3,6][1,-3] / magnitude of [-3,6] x magnitude of [1,-3]

= -3 - 18 / square root of 9+36 x square root of 1+9

= -21 / square root of 450

I didn't even calculate the angle between B and C, and i'm already over 180 degrees so where did I go wrong?

2. Consider the three distances $\displaystyle AB,~AC,~\&~,BC$.
Do they form a Pythagorean Triple?

3. But the 2 angles I calculated go over 180 degrees (what a triangle is made up of by all of its angles), so where did I go wrong in my calculations?

4. You did not go wrong. You just find the other angle which is bitween AB and BC, and that is the angle 171.9. The angle that you need is 180-171.9=28.1.

Just draw the points on coordinate system, and see what angles you need.

Regards.

5. But the angle between A and B is 85.2 degrees + the angle between B and C is 171.9 degrees = 257.1 degrees

Am I missing something? 257.1 degrees is way over 180 degrees Don't I have to add all 3 angles? AB + BC + AC = 180 degrees.. So far I only calculated AB and BC and i'm at 257.1 degrees.

6. Originally Posted by kmjt
Given A(5,2), B(-3,6), and C(1,-3), determine if triangle ABC is a right triangle.

So first I took a look at the angle between A and B:

cos theta = [5,2][-3,6] / magnitude of [5,2] x magnitude of [-3,6]

= -15 + 12 / square root of 25+4 x square root of 9+36

= -3 / square root of 1305

Then I took a look at the angle between B and C:

cos theta = [-3,6][1,-3] / magnitude of [-3,6] x magnitude of [1,-3]

= -3 - 18 / square root of 9+36 x square root of 1+9

= -21 / square root of 450

I didn't even calculate the angle between B and C, and i'm already over 180 degrees so where did I go wrong?

You should consider the angle between BA and BC and so on. You are calculating the angle whose vertex is at the origin and subtending AB.

7. you do not calculate the angle between two points ... you calculate the angle between two vectors.

the angle between vector AB and vector AC is angle A.

AB = $\displaystyle \left<-8,4\right>$

AC = $\displaystyle \left<-4,-5\right>$

$\displaystyle A = \arccos\left(\frac{AB \cdot AC}{|AB| |AC|}\right)$

$\displaystyle A = \arccos\left(\frac{12}{\sqrt{3280}}\right) = 77.9^\circ$

8. Taking Plato's advice, i.e. checking whether the points A, B, and C are the vertices of a right-angled triangle, is probably more neat.