# Need help reviewing a trig question for exam

• Jun 18th 2008, 06:31 AM
mathdonkey
Need help reviewing a trig question for exam
I'm studying for an exam and I could use a reminder on how to solve this.

"A right triangle with a 90 degree angle on the bottom right has a hypotenuse that is 50 units long. The bottom left angle is a 48 degree angle. Solve for the side between the right angle and the reference angle correct to one decimal point."
• Jun 18th 2008, 07:23 AM
Isomorphism
Quote:

Originally Posted by mathdonkey
I'm studying for an exam and I could use a reminder on how to solve this.

"A right triangle with a 90 degree angle on the bottom right has a hypotenuse that is 50 units long. The bottom left angle is a 48 degree angle. Solve for the side between the right angle and the reference angle correct to one decimal point."

Actually the question is asking if you can calculate $\cos 48 ^{\circ}$ upto 2 decimal places?

If you draw the triangle, the required side is clearly $50\cos 48^{\circ}$
• Jun 18th 2008, 07:49 AM
mathdonkey
Quote:

Originally Posted by Isomorphism
Actually the question is asking if you can calculate $\cos 48 ^{\circ}$ upto 2 decimal places?

If you draw the triangle, the required side is clearly $50\cos 48^{\circ}$

Thanks I appreciate the help.

I don't want the answer for this question so much as the process for solving it (what to do and why). My notes on this aren't very clear. Could you help me out?
• Jun 18th 2008, 11:12 AM
masters
Quote:

Originally Posted by mathdonkey
Thanks I appreciate the help.

I don't want the answer for this question so much as the process for solving it (what to do and why). My notes on this aren't very clear. Could you help me out?

$\cos\angle{A}=\frac{adjacent \ \ side}{hypotenuse}$

$\sin\angle{A}=\frac{opposite \ \ side}{hypotenuse}$

$\tan\angle{A}=\frac{opposite \ \ side}{adjacent \ \ side}$