# trig problem

• Sep 23rd 2008, 10:55 PM
erico
trig problem
i have a triangle with all angles and one side given. i need to find the other sides but dont know the steps to do so
• Sep 23rd 2008, 10:59 PM
Chop Suey
Use law of sines. Given a triangle ABC, side a is opposite angle A, side b is opposite angle B, side c is opposite angle C, then Law of Sines state that:

$\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}$
• Sep 23rd 2008, 11:03 PM
erico
so if i have angles 38, 55, and 87 degrees and angle 38 has the given side of 500 meters opposite how would i set it up?
• Sep 23rd 2008, 11:13 PM
Chop Suey
$\frac{\sin{38}}{500} = \frac{\sin{55}}{x}$

Solving for x yields the value of the side opposite angle 55.
• Sep 23rd 2008, 11:22 PM
erico
ok i got answers of 665.26 and 811.02 but when i use path. theorm it doesnt work out. did i do somethin wrong?
• Sep 23rd 2008, 11:24 PM
Chop Suey
Quote:

Originally Posted by erico
ok i got answers of 665.26 and 811.02 but when i use path. theorm it doesnt work out. did i do somethin wrong?

Pythagorean theorem works only for right-angle triangles.
• Sep 23rd 2008, 11:34 PM
erico
ok thank you very much. i also had some trouble with another problem. its given sides are BC 160, AD 270, and DC 190. it also has angles ADC 75 degrees and angle BCD is 90 degrees. i started by tryin to cut it to make two triangles. my separation was from angle B to D. which left me with triangles ABD and BCD. did i start off correctly?
• Sep 24th 2008, 12:38 AM
Patrick Murray
The Sine Rule
For a great animated and narrated explanation of this, check out

Mathematics.com.au - The Sine Rule

You will understand it easily.