# Thread: Stuck on this trig problem.

1. ## Stuck on this trig problem.

Hey, here is the problem.

In triangle ABC, side "a" is twice as long as side "b". If the hypotenuse is 30cm long, solve the triangle.

If you could please explain to me how to solve this, it would be much appreciated.

2. ## Re: Stuck on this trig problem.

Let side 'b' be $x$cm

Then, label each side of the triangle with respect to $x$.

Then use Pythagoras' theorem.

Edit: It will be easier to let side b be $x$cm, rather than side a, as I had suggested initially.

3. ## Re: Stuck on this trig problem.

Originally Posted by Quacky
Let side 'b' be $x$cm

Then, label each side of the triangle with respect to $x$.

Then use Pythagoras' theorem.

Edit: It will be easier to let side b be $x$cm, rather than side a, as I had suggested initially.
ok, so if side b is "x" than side a would be "2x"? what would side "c" be then?

4. ## Re: Stuck on this trig problem.

"The hypotenuse is 30cm long"

5. ## Re: Stuck on this trig problem.

ok so would the equation to solve this value be:?

$x^2 + 2x^2 = 30cm^2$

and if so, how do I find the value of X?
sorry for all the questions, this question doesn't seem like it should he hard, but its giving me trouble for some reason.

6. ## Re: Stuck on this trig problem.

If you mean $x^2+(2x)^2=30^2$, where x is in cm, then yeah.

Edit: Error corrected.

7. ## Re: Stuck on this trig problem.

Originally Posted by Quacky
If you mean $x^2+2x^2=30^2$, where x is in cm, then yeah.
I've been trying to find the value of X here but am having a hard time. I already know the answer to my original problem is a=26.82cm and b=13.41cm, but I have no idea on how to derive that from this formula.

8. ## Re: Stuck on this trig problem.

In triangle ABC, side "a" is twice as long as side "b". If the hypotenuse is 30cm long, solve the triangle.
$a = 2b$

$a^2 + b^2 = 30^2$

$(2b)^2 + b^2 = 30^2$

$5b^2 = 30^2$

$b^2 = \frac{30^2}{5}$

$b = \frac{30}{\sqrt{5}} = 6\sqrt{5}$

the sides are ...

$a = 12\sqrt{5}$

$b = 6\sqrt{5}$

$c = 30$

9. ## Re: Stuck on this trig problem.

Originally Posted by skeeter
$a = 2b$

$a^2 + b^2 = 30^2$

$(2b)^2 + b^2 = 30^2$

$5b^2 = 30^2$

$b^2 = \frac{30^2}{5}$

$b = \frac{30}{\sqrt{5}} = 6\sqrt{5}$

the sides are ...

$a = 12\sqrt{5}$

$b = 6\sqrt{5}$

$c = 30$
thanks alot, I was getting that last step wrong, makes sense now.

10. ## Re: Stuck on this trig problem.

I'd made an error in my original calculation, which didn't help either.