Step by step
The question
Question: find side b of the triangle with a = 7, c = 9 and B = 130°.
1 Write the law of cosines
We know two sides, a and c, and the angle B between them.
The law of cosines: b² = a² + c² − 2ac cos B.
Side b is across from angle B.
It is Pythagoras with a correction.
When B is 90°, cos 90° is 0, and it is just Pythagoras.
2 Put the numbers in
Now put the numbers in: a = 7, c = 9 and B = 130°.
3 Work it out
Squares first: 7² is 49, 9² is 81.
And 2 × 7 × 9 is 126.
Careful!
Times comes before take away.
Work out 126 × cos 130° first, then take it away.
The calculator gives cos 130°: about −0.6428.
126 times that is about −80.99.
49 + 81 is 130.
Then we take away −80.99.
Taking away a negative is adding: − (−80.99) = + 80.99.
The angle is more than 90°, so b is longer than Pythagoras gives.
So b² ≈ 210.99.
4 Take the square root
b squared is 210.99.
So b is the square root of 210.99.
A side is never negative, so only the plus root.
The calculator gives about 14.525.
To the nearest tenth, b ≈ 14.5.
5 Check
Let's check: the law turned around gives cos B back.
130 take away 210.99, over 126: −0.6428.
And cos 130° is about −0.6428.
The same!
So b 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.