Step by step
The question
Question: find side c of the triangle with a = 4, b = 7 and C = 110°.
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 = 4, b = 7 and C = 110°.
3 Work it out
Squares first: 4² is 16, 7² is 49.
And 2 × 4 × 7 is 56.
Careful!
Times comes before take away.
Work out 56 × cos 110° first, then take it away.
The calculator gives cos 110°: about −0.3420.
56 times that is about −19.15.
16 + 49 is 65.
Then we take away −19.15.
Taking away a negative is adding: − (−19.15) = + 19.15.
The angle is more than 90°, so c is longer than Pythagoras gives.
So c² ≈ 84.15.
4 Take the square root
c squared is 84.15.
So c is the square root of 84.15.
A side is never negative, so only the plus root.
The calculator gives about 9.173.
To the nearest tenth, c ≈ 9.2.
5 Check
Let's check: the law turned around gives cos C back.
65 take away 84.15, over 56: −0.342.
And cos 110° is about −0.3420.
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.