# finding trigonometry of angles

• Mar 15th 2007, 04:17 PM
zachary7
finding trigonometry of angles
I was given an assignment and I really need some help. I've not been doing algebra in the other school and have no idea how to do applications and concepts. The paper is special right triangles and I must "find each missing length. Round to the nearest tenth if necessary." My question is HOW do you solve these triangle problems?????????
Many thanks to anyone who can offer a hand.
Zach
• Mar 15th 2007, 06:46 PM
Aryth
Quote:

Originally Posted by zachary7
I was given an assignment and I really need some help. I've not been doing algebra in the other school and have no idea how to do applications and concepts. The paper is special right triangles and I must "find each missing length. Round to the nearest tenth if necessary." My question is HOW do you solve these triangle problems?????????
Many thanks to anyone who can offer a hand.
Zach

Well, there are two particular special right triangles:

One is the 30,60,90, which are the angles of the triangles:

http://www.mathnstuff.com/math/spoke...f/a306090n.gif

Where x = 1, y = sqrt(3), and r = 2 r is the resultant or the hypotenuse as you'll see it.

The other special triangle is the 45-45-90 triangle:

http://www.mathnstuff.com/math/spoke...if/a4545no.gif
Where x = 1, y = 1, and r = sqrt(2)

For any other right triangles, you can use these formulas to solve for x,y,r and the angles:

cos(@) = x/r
sin(@) = y/r
tan(@) = y/x

If r = 1, then:

cos(@) = x
sin(@) = y
tan(@) = y/x

Or if you have just the lengths, you can use this:

x^2 + y^2 = r^2

This is called the Pythagorean Theorem.

Where @ is the angle in degrees or radians, x is the horizontal component, otherwise known as the bottom leg, y is the vertical component, known as the upper leg, and r is the hypotenuse, it is always the side that slants from one leg to another. I hope this helps.