# Thread: Very basic trig problem

1. ## Very basic trig problem

I understand that $c^2 = a^2+b^2$

Now I have this problem, find the missing a,b or c length.

$a=\sqrt{2}, b=\sqrt{2}$

So I say that $a^2=4, b^2=4, c^2=8$

or am I doing it wrong?

If not then I get confused on this next question.

$a=\sqrt{8}, c=3$

$a^2=64, c^2=9$ but how can $c^2 = a^2+b^2$ in this question?

2. ## Re: Very basic trig problem

The square of the square root of 2 is 2, not four.

The square of the square root of 8 is 8, not 64.

3. ## Re: Very basic trig problem

I know but if the square root of 8 is 8, then $8^2$ must equal 64. Isn't the square root the opposite of squared?

4. ## Re: Very basic trig problem

The square root of 8 is not 8. The square root of 8 is 2 times the square root of two.

You seem to be reading "a = square root of 8" as "a squared = 8".

In you "a = " statements, try squaring both sides. Maybe that'll make it clearer.

5. ## Re: Very basic trig problem

This academic difficult for me really good

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