Step by step
The question
Question: find side c of the triangle with a = 9, b = 12 and C = 38°.
1 Write the law of cosines
We know two sides, a and b, and the angle C between them.
The law of cosines: c² = a² + b² − 2ab cos C.
Side c is across from angle C.
Where does it come from?
Put corner C at 0, with side a along the x-axis.
The last corner is b times the unit-circle point: b cos C across, b sin C up.
Side c joins two corners.
Pythagoras on the across and up gaps gives c squared.
Open the brackets.
Take b² out of the two square parts.
cos squared plus sin squared is 1.
That leaves the law.
When C is 90°, cos 90° is 0, and it is just Pythagoras.
2 Put the numbers in
Now put the numbers in: a = 9, b = 12 and C = 38°.
3 Work it out
Squares first: 9² is 81, 12² is 144.
And 2 × 9 × 12 is 216.
Careful!
Times comes before take away.
Work out 216 × cos 38° first, then take it away.
The calculator gives cos 38°: about 0.7880.
216 times that is about 170.21.
81 + 144 is 225.
So c² ≈ 54.79.
4 Take the square root
c squared is 54.79.
So c is the square root of 54.79.
A side is never negative, so only the plus root.
The calculator gives about 7.402.
To the nearest tenth, c ≈ 7.4.
5 Check
Let's check: the law turned around gives cos C back.
225 take away 54.79, over 216: 0.788.
And cos 38° is about 0.7880.
The same!
So c is right.
Let's look back.
First: write the law of cosines.
Then: put the numbers in.
Then: work it out.
Then: take the square root.
Last: check.