anonymous
  • anonymous
Trigonometric functions Help! I'm lost here...
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
Luigi0210
  • Luigi0210
Well you are given two lengths, why not use tanA?
Luigi0210
  • Luigi0210
TanA=4/5

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
because that's not on the multiple choice, or is it? I seriously do not understand these principles...
Luigi0210
  • Luigi0210
reason being because the problem is not finished \[tanA=4/5\] \[\tan^{-1} (tanA)=\tan^{-1} (4/5)\] \[A=\tan^{-1} 4/5\] Plug that into a calculator and that should give you your answer
anonymous
  • anonymous
ok so this one is tan^(-1) 5/7, right?
1 Attachment
Luigi0210
  • Luigi0210
correct
anonymous
  • anonymous
ok so they switched this one up on me. so would it be tan^(-1) 5/4?
1 Attachment
Luigi0210
  • Luigi0210
Correct again
anonymous
  • anonymous
YES!!! *happy dance* thank you so much. You have no idea how long i have been trying to figure these things out. *e-hug*
Luigi0210
  • Luigi0210
ha, you're welcome ^_^'
anonymous
  • anonymous
So what about this one?
1 Attachment
Luigi0210
  • Luigi0210
Use sin
anonymous
  • anonymous
Thanks! I worked that one out, do you know how to do these? I'm stuck on this one.
1 Attachment
Luigi0210
  • Luigi0210
it's a sinx graph: 2sin(x+.5Pi)+1
anonymous
  • anonymous
ok, what about this one?
1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.