Let A= (1,3), B=(6,0), and C= (9,9). Find the size of angle BAC.

Question

Let A= (1,3), B=(6,0), and C= (9,9). Find the size of angle BAC.

Hint: add x=1 into the diagram and think about how to draw in right triangles to solve the problem.

Answer 1

Draw a sketch. One way to do this is to make a lower and upper right triangle like this:

The lower right triangle has an angle α at the top and the upper right triangle has an angle β at the bottom. ∠α + ∠β + ∠BAC = 180°

You can find ∠α using the trig ratio tangent. tan(α) = 5/3 or α = tan^-1(5/3)

You can find ∠β using the trig ratio tangent. tan(β) = 8/6 or α = tan^-1(8/6)

Then using ∠α + ∠β + ∠BAC = 180°, you can calculate ∠BAC:

∠BAC = 180° - ∠α - ∠β

Answer 2

Law of Cosines:

a=sqr90, b=sqr34, c = 10

a^2 = b^2 + c^2 -2bcCosA

90 =34+100 -2(10)sqr(34)CosA

-44 = -20sqr34CosA

CosA = 44/20sqr34

= 11/5sqr34

A = Cos^-1(11/5sqr34) or Arccos(11/5sqr34)

= 67.83365418

= about 68 degrees

= about 67.83(pi/18) radians

=about .38pi radians

= about 1.184 radians

A(1,3) B(6,0) C(9,9)

angle ABC = angle B = about 77.47 degrees

from A to B = sqr(5^2+3^2) = sqr34=c = side opposite angle C

B to C =sqr(9^2+3^3) = sqr90= 3sqr10=a = side opposite angle A

A to C = sqr(8^2+6^2) = sqr100=10=b = side opposite angle B

use the law of cosines

b^2 = a^2 +c^2 -2acCosB

100 = 90+34 -2(3)sqr(34)(10))CosB

100-124 =-6sqr340CosB

-24 = -6sqr340CosB

4 = sqr340CosB

CosB = 4/sqr340

B = angle ABC = Cos^-1(4/sqr340) = about 77.4712 degrees

it's not that far from a right triangle or 90 degrees for B

or

B = 180- tan^-1(3) - tan^-1(3/5)

=about 180-71.57-31 = 180-102.57

= about 77.33 degrees with more rounding

go with 77.47 from the Law of Cosines, the standard way to find an angle when you know 3 sides

slope of AB is sort of almost the negative inverse of the BC slope

-3/5 = AB slope= -9/15=-27/45

9/3 = 3 = BC slope= 45/15

if you roughly sketch or graph the 3 points, connect them, it visually looks a little less than 90 degrees

77.47 degrees =

77.47pi/180 radians =

about .43pi radians

= about 1.352 radians

C= 180-67.83-77.47

= 180-145.3

= 34.7 degrees

angle ACB = BCA = about 37.4 degrees

= about .2pi radians

= about .653 radians

Answer 3

Vector AB = <6 - 1, 0 - 3> = <5, -3>; vector AC = <9 - 1, 9 - 3> = <8, 6>

cos(BAC) = (5·8 - 3·6)/[(25 + 9)(64 + 36)]^1/2 = 22/(10√34) = 11√34/170; BAC = cos^-1(11√34/170) = 67.834°