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I solved it up till 30 degrees but then they're saying the "since AC is the largest side" I don't get it
do you remember your unit circle?
So do you remember that when we have sine of an angle, it can have two values?
yea you minus 30 180 yea I know that but visually I just can't imagine it in this triangle
30 minus from 180
yeah, that's how you get 150 degrees... If theta = 30, it means that angle ABC and angle ACB would be larger than 30 true? Since the ANGLE SUM of a triangle = 180 right? If |dw:1434891872359:dw|
wait so what I'm understanding from this that sum of angle ABC and ACB is 150??
not quite... Just remember that the angle that corresponds to the longest side of a triangle has the largest angle
Ok I still don't 100% get it. Okay like suppose I don't that there was possibly another length of AC that is longer. compare to when it is if you have our theta as 30 degrees. How do you prove that the other angle of theta would be 150 by calculation??
|dw:1434892380598:dw| So when dealing with sine there are two values... but to test it, you consider the ambigous case. So if you have the other angles, you can add them all up and make sure that they equal to 180 (since the angle sum of a triangle = 180)
is this what you're saying..?
This is what's given in my text book ofc I didn't understand it first but after some while I pictured it to be in that way as you drew in the unit circle diagram. The thing is trying to put our triangle into this unit cirle I can't imagine how I'd do that..
lets see |dw:1434892824826:dw|
uhh.. you mean like this right? |dw:1434892995285:dw|
oh no nvm, they showed you the formula using area.
I, though, haven't seen the formula using area.
I know the answer sir And the know the methods used to find 30 what I'm looking for is the proof for 150 degrees
30 can not be the largest angle of any triangle. The sum of 3 angles must be 180.
if 30 is the largest, the other 2 angles are less than 30 and their sum is max 90...
@hartnn Wait wait lets not go there. can we just stick to the unit circle? it's much easier for me to understand. Look cause I'm trying to put the triangle from our original question into the unit circle. |dw:1434894775692:dw| I just wanna know if I'm visually imagining it correctly. Thats all
yes thats correct. so now label side AC
|dw:1434895631747:dw| @hartnn Thanks a BUNCH!
correct :) welcome ^_^