# Problem with IB past paper, Triangle and sin

• Mar 15th 2010, 06:37 AM
Leitzge
Problem with IB past paper, NOT SOLVED
Hello everyone, I do Math SL for the IB, and I don't fully understand the following question that I've attached as an image to this post. I would appreciate any kind of help.

I know that sin of th. is the opposite / hypotenuse, and the hypotenuse is 3, but why is the opposite sqrt. of 5?

also question b I don't full understand.
• Mar 15th 2010, 07:17 AM
Sudharaka
Quote:

Originally Posted by Leitzge
Hello everyone, I do Math SL for the IB, and I don't fully understand the following question that I've attached as an image to this post. I would appreciate any kind of help.

I know that sin of th. is the opposite / hypotenuse, and the hypotenuse is 3, but why is the opposite sqrt. of 5?

also question b I don't full understand.

Dear Leitzge,

If you apply the Phytogoras theorem to the traingle,

$AB^{2}=AC^{2}+CB^{2}$

$CB^{2}=AB^{2}-AC^{2}$

$BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{9-4}=\sqrt{5}$

Therefore, $sin\theta=\frac{\sqrt{5}}{3}$

Similarly you could calculate $cos\theta$ then use,

$sin2\theta=2\sin{\theta}\cos{\theta}$

I have told you how to get the answer. All you got to do is calculate $sin\theta~and~ cos\theta$ by looking at the traingle. Then you have to substitute those values in the given equation, $sin2\theta=2\sin{\theta}\cos{\theta}$