Step by step
The question
Question: find side c of the triangle with a = 6 m, b = 9 m and C = 100°.
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.
It is Pythagoras with a correction.
When C is 90°, cos 90° is 0, and it is just Pythagoras.
2 Put the numbers in
Now put the numbers in: a = 6, b = 9 and C = 100°.
3 Work it out
Squares first: 6² is 36, 9² is 81.
And 2 × 6 × 9 is 108.
Careful!
Times comes before take away.
Work out 108 × cos 100° first, then take it away.
The calculator gives cos 100°: about −0.1736.
108 times that is about −18.75.
36 + 81 is 117.
Then we take away −18.75.
Taking away a negative is adding: − (−18.75) = + 18.75.
The angle is more than 90°, so c is longer than Pythagoras gives.
So c² ≈ 135.75.
4 Take the square root
c squared is 135.75.
So c is the square root of 135.75.
A side is never negative, so only the plus root.
The calculator gives about 11.651.
To the nearest tenth, c ≈ 11.7 m.
5 Check
Let's check: the law turned around gives cos C back.
117 take away 135.75, over 108: −0.1736.
And cos 100° is about −0.1736.
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.