#finding other side of right triangle
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...Pythagorean theorem.
Wait, I'm confused about what you're asking.
but i need to find a
Remember SOHCAHTOA, right?
Do you remember SOHCAHTOA?
the trig identities?
What do you think I'm talking about?
i mean thats the only thing i could think of lmao
ok so its an identity thing left
Right, but what specifically are you thinking about?
how they are reciprocals of each other
What are reciprocals of what?
What is "it"?
sin A
cos A
That's not an "it", that's a "them".
oh
Do you remember SOHCAHTOA or not?
if you are talking about like cos = adj/hyp then yes
Okay. So what side is opposite A?
yeah but im not solveing for cosecant
Will you answer the question I asked?
...yes, that's the question I asked. Are you going to answer it?
It's a side.
In the picture, what side is opposite A?
yes
Frankly, I don't even understand what you were talking about half the time and I suspect you kind of didn't either. Like, why did you think anything I was saying was talking about the cosecant of anything?
i was thinking that if we inverse sin itll automatically equal cosecant
since cosecant is literally the reciprocal of sin
yeah, but cosine and cosecant are different things
btw guys i apologize if i actually sound stupid, im taking trig this semester without having taken geometry but having done very well in precalc
so i thought i would do better
i dont think im doing bad since i literally started like 3 days ago but still
i think you’re doing fine
tbh if you could solve it correctly for the blue angle i would’ve suggested you just redo the process for the red angle
yeah
but knowing sin(A) = cos(B) is always handy
i realized (i think) that its literally just reversed since two angles must equal 180
(B is blue angle)
so if we know one is 90 we know the other is literally 90
they add up to 90
i think
oh yeah my fault
ur right
but i messed up in the beginning by labeling them weird anyways
i shouldve labeled A as B and B as A
The reason it's called the cosine is because it's the "sine" of the "co"mplementary angle.
huh
so i got
SinA = 19/sqrt410
CosA = 7/sqrt410
sin(90 - x) = cos(x)
cos(90 - x) = sin(x)```(in degrees)
which is opposite of what i got the first time which was
SinA = 7/sqrt410
CosA = 19/sqrt410
you labelled A and B correctly, you just somehow solved for sin(B) and cos(B)