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How do I find the length of a side of a triangle using the cosine rule?

Suppose we have a triangle with 2 sides of length 4 and 5 (in any units), and the angle between those sides is 60 degrees. We want to find the length of the side opposite the 60 degree angle.

Write the cosine rule, a2=b2+c2-2bcCos(A). Here a, b and c are the side lengths, and A is the angle opposite side a.

Given our example we have a2=42+52-2*4*5*Cos(60)

so a2=16+25-20 = 21, a = sqrt(21) which is irrational (approximately 4.58).

We can also use the cosine rule if we know two side lengths and an angle opposite one of those sides. Suppose we have a triangle with side of length 2 and 5, and we know that the angle opposite the side of length 5 is 120 degrees. We want to work out the 3rd side length.

Again we write out a2=b2+c2-2bcCos(A), but this time we want a to be the side of length 5, since this side is opposite the known angle, giving us

52=22+c2-2*5*c*Cos(120)

25=4+c2-10*c*(-0.5)

c2+5c-21=0

Solving with the quadratic formula we get

c = (-5 +/- sqrt(25+84)) / 2 which is again irrational.

Notice this time we get two different values for c, since in this case there are two possible triangles that meet the criteria described in the question.

Angus H. GCSE Maths tutor, A Level Maths tutor, GCSE Physics tutor, A...

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