Step by step
The question
Question: find side c of the triangle with a = 8, b = 11 and C = 47°.
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 = 8, b = 11 and C = 47°.
3 Work it out
Squares first: 8² is 64, 11² is 121.
And 2 × 8 × 11 is 176.
Careful!
Times comes before take away.
Work out 176 × cos 47° first, then take it away.
The calculator gives cos 47°: about 0.6820.
176 times that is about 120.03.
64 + 121 is 185.
So c² ≈ 64.97.
4 Take the square root
c squared is 64.97.
So c is the square root of 64.97.
A side is never negative, so only the plus root.
The calculator gives about 8.06.
To the nearest tenth, c ≈ 8.1.
5 Check
Let's check: the law turned around gives cos C back.
185 take away 64.97, over 176: 0.682.
And cos 47° is about 0.6820.
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.