So I just need help starting this please.. Part 1: Make up an angle measure for angle C and a length for one of the sides (AB, BC, or CA). Part 2: Use a trigonometric ratio (sine, cosine, or tangent) to solve for one of the other side lengths.
Give me an angle between say 25 and 65
and any length for the side keep it simple like a whole number
So 50 degrees then like 35? o.o
good, lets use sine. If the angle is for c then the opposite is the section BA and the Adjacent is CB. Sin 50 = .766 and the length of BA is the 35 you set. So .766=35/CB
solve for CB and you have a length for another side as asked. you can then get the Hypotenuse by taking the square root of the sum of the squares for CB and BA, but that wasn't asked for...
CB=35/.766 =45.689 follow the thought process I laid out and make sure I didn't make a mistake.
I did make a mistake
sine is opposite over hypotenuse, not adjacent... something seemed off. so you are solving for CA there, not BC...
I'm confused:/
Let me clarify, sorry to make the error... Sine = opposite / Hypotenuse. The opposite is the bottom and the Hypotoenuse is the longest side (CA). taking sin(50) = .766 so since sin(50) = opposite / hypotenuse you can replace sin(50) with .766. That gives you .766=opposite/Hypotenuse. You assigned the length of the opposite to be 35 so your equation becomes .766 = 35/hypotenuse. you can solve for the length of the hypotenuse by multiplying each side by the hypotenuse and dividing each side by .766. This gives you: Hypotenuse = 35/.766. The answer to this is 45.689
woah, this is some complicated stuff but it makes more sense now in a way.
It helps when then person explaining it doesn't get it backwards. It also helps to print out a cue sheet that physically shows what is going on. There is a really good one at: http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf
Thank yo so much!! I'm gonna save that link to my favs.
Join our real-time social learning platform and learn together with your friends!