Step by step
The question
Question: find side a of the triangle with b = 6, c = 10 and A = 90°.
1 Write the law of cosines
We know two sides, b and c, and the angle A between them.
The law of cosines: a² = b² + c² − 2bc cos A.
Side a is across from angle A.
It is Pythagoras with a correction.
When A is 90°, cos 90° is 0, and it is just Pythagoras.
2 Put the numbers in
Now put the numbers in: b = 6, c = 10 and A = 90°.
3 Work it out
Squares first: 6² is 36, 10² is 100.
And 2 × 6 × 10 is 120.
Careful!
Times comes before take away.
Work out 120 × cos 90° first, then take it away.
The calculator gives cos 90°: exactly 0.
120 times that is 0.
36 + 100 is 136.
So a² = 136.
4 Take the square root
a squared is 136.
So a is the square root of 136.
A side is never negative, so only the plus root.
The calculator gives about 11.662.
To the nearest tenth, a ≈ 11.7.
5 Check
Let's check: the law turned around gives cos A back.
136 take away 136, over 120: 0.
And cos 90° is 0.
The same!
So a 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.