Step by step
The question
Question: find side b of the triangle with a = 9, c = 4 and B = 50°.
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 = 9, c = 4 and B = 50°.
3 Work it out
Squares first: 9² is 81, 4² is 16.
And 2 × 9 × 4 is 72.
Careful!
Times comes before take away.
Work out 72 × cos 50° first, then take it away.
The calculator gives cos 50°: about 0.6428.
72 times that is about 46.28.
81 + 16 is 97.
So b² ≈ 50.72.
4 Take the square root
b squared is 50.72.
So b is the square root of 50.72.
A side is never negative, so only the plus root.
The calculator gives about 7.122.
To the nearest tenth, b ≈ 7.1.
5 Check
Let's check: the law turned around gives cos B back.
97 take away 50.72, over 72: 0.6428.
And cos 50° 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.