# Had a test today.. Need to know if this is right.

• Nov 4th 2010, 06:13 PM
veryverymuch
Had a test today.. Need to know if this is right.
There's a right triangle with hypotenuse 5. I forgot what term it used, but they said something about it being in between 0 and $\frac{\pi}{2}$

x is adjacent to theta. Find $sin2\theta$ I already know that $sin2\theta=2sin\theta cos\theta$, it's just the length of the sides I was having trouble with.

Am I correct in assuming this is just your standard 3, 4, 5 triangle and the $\theta = \frac{24}{25}$

There was another way I tried to solve it.. where y(opposite to theta) was = $\sqrt{25-x^2}$ but I didn't think that was correct so I went with the assumption I mentioned above..
• Nov 4th 2010, 06:17 PM
mr fantastic
Quote:

Originally Posted by veryverymuch
There's a right triangle with hypotenuse 5. I forgot what term it used, but they said something about it being in between 0 and $\frac{\pi}{2}$

x is adjacent to theta. Find $sin2\theta$ I already know that $sin2\theta=2sin\theta cos\theta$, it's just the length of the sides I was having trouble with.

Am I correct in assuming this is just your standard 3, 4, 5 triangle and the $\theta = \frac{24}{25}$

There was another way I tried to solve it.. where y(opposite to theta) was = $\sqrt{25-x^2}$ but I didn't think that was correct so I went with the assumption I mentioned above..

The sine and cosine of an angle in a 3-4-5 triangle is pretty simple to get, I would have thought. As is the substitution of the values into the double angle formula. If you say what answer you gave, someone can confirm or reject it.
• Nov 4th 2010, 06:20 PM
veryverymuch
Quote:

Originally Posted by mr fantastic
The sine and cosine of an angle in a 3-4-5 triangle is pretty simple to get, I would have thought. As is the substitution of the values into the double angle formula. If you say what answer you gave, someone can confirm or reject it.

Sorry if I was unclear. The details the question gave:
5 is the hypotenuse
It is a right triangle.

The answer I got was $\theta=\frac{24}{25}$ when solving $sin2\theta$

What I want to know is if it WAS a 3 4 5 triangle even though the question didn't state it. The wording is the only thing that confused me.
• Nov 4th 2010, 08:28 PM
mr fantastic
Quote:

Originally Posted by veryverymuch
Sorry if I was unclear. The details the question gave:
5 is the hypotenuse
It is a right triangle.

The answer I got was $\theta=\frac{24}{25}$ when solving $sin2\theta$

What I want to know is if it WAS a 3 4 5 triangle even though the question didn't state it. The wording is the only thing that confused me.

I think you will find that there are not too many right-triangles with a hypotenuse of length 5 other than a 3-4-5 triangle. I suggest you research Pythagorean triples ....

By the way, $\theta=\frac{24}{25}$ is not a correct statement. Surely you mean $\sin(2\theta)=\frac{24}{25}$.
• Nov 4th 2010, 11:51 PM
veryverymuch
Quote:

Originally Posted by mr fantastic
I think you will find that there are not too many right-triangles with a hypotenuse of length 5 other than a 3-4-5 triangle. I suggest you research Pythagorean triples ....

By the way, $\theta=\frac{24}{25}$ is not a correct statement. Surely you mean $\sin(2\theta)=\frac{24}{25}$.

Yeah, I meant $sin2\theta$
Thanks either way, though.