Point A = (-4,3) Point B = (2,1) Point C = (-2,-3) Find the size of angle BAC?

Question

I know that the length of AC and AB is sqrt40 and the length of BC is sqrt32. Where do I go from here?

Answer 1

If you draw out the triangle on a Cartesian coordinate plane

you can see where the points fall in relation to each other.

 

The law of cosines says:

c² = a² + b² - (2ab)Cos C

 

In this case, C is going to represent our angle BAC

So let's call this

c² = a² + b² - (2ab)Cos(BAC)

 

c is the length of the side opposite our angle which is √32 from the

length of line segment BC as you already determined.

 

a and b are the lengths of the adjacent sides which are both √40

from the lengths of line segments AC and AB as you determined.

 

(√32)² = (√40)² + (√40)² -2(√40)(√40)Cos(BAC)

 

32 = 40 + 40 -80 Cos(BAC)

 

32 = 80 - 80Cos(BAC)

 

80Cos(BAC) = 48

 

Cos(BAC) = 48/80

 

Now we use the inverse cosine function Cos^-1 to find

the measure of angle BAC

 

BAC = Cos^-1 (48/80)

 

BAC = 53.13^o

 

 

Answer 2

Hi Emmalie.

 

We can first work out the lengths AB, AC and BC. We can then use the law of cosines to work out angle BAC, which I will call a for now.

 

You can work out AB as sqrt[(2 - (-4))² + (1 - 3)²] = sqrt[6² + (-2)²] = sqrt[36 + 4] = sqrt[40]

 

AC = sqrt[(-2 - (-4))² + (-3 - 3)²] = sqrt[2² + (-6)²] = sqrt[40]

 

BC = sqrt[(-2 - 2)² + (-3 - 1)²] = sqrt[(-4)² + (-4)²] = sqrt[32] 

 

It would help to draw a picture of this triangle. You want the angle that is opposite to the side BC. So, the Law of Cosines says

 

(BC)² = (AC)² + (AB)² - 2(AC)(AB)cos(a)

 

Or

 

(√32)² = (√40)² + ( √40)² - 2 (√40)(√40)cos(a)

 

Or

 

(√32)² = (√40)² + ( √40)² - 2 (√40)²cos(a)

 

Or

 

32 = 40 + 40 -(2)(40)cos(a)

 

32 = 80 - 80cos(a)

 

Divide by 8.

 

4 = 10 - 10cos(a)

 

-10cos(a) = -6

 

cos(a) = 3/5

 

a = arcos(0.6)

 

a = 53.13 deg