If I am give the the sides of a triangle, where
Side RA=4
Side AB=4.8
Side RB=7.1
How do I find Angle RAB?
Thanks!
Law of Cosines
C^2=a^2+b^2-2abCosC
so label your sides a, b, c
you want angle c to equal RAB
so RB=C
RA=a
AB=b
To find Angle RAB, you can use the Law of Cosines, which states that the square of one side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of those two sides and the cosine of the angle between them.
In this case, you are given the lengths of all three sides of the triangle (RA = 4, AB = 4.8, and RB = 7.1), and you want to find Angle RAB.
Using the Law of Cosines, you can set up the equation as follows:
RA^2 = AB^2 + RB^2 - 2 * AB * RB * cos(RAB)
Substituting the known values, you get:
4^2 = 4.8^2 + 7.1^2 - 2 * 4.8 * 7.1 * cos(RAB)
16 = 23.04 + 50.41 - 67.68 * cos(RAB)
Rearranging the equation:
-57.45 = -67.68 * cos(RAB)
Now, divide both sides by -67.68:
cos(RAB) = -57.45 / -67.68
cos(RAB) ≈ 0.8494
To find the actual angle RAB, you need to use the inverse cosine function (cos^(-1)) or arccosine. Using a scientific calculator or online tool, you can find that the inverse cosine of 0.8494 is approximately 31.46 degrees.
Therefore, Angle RAB is approximately 31.46 degrees.