anonymous
  • anonymous
Please Help Will Medal And Fan! For standard position angle determined by the point (x,y) what are the values of the trigonometric functions. For the point (5,12) find csc theta and sec theta
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
A. Csc theta= 13/12 , sec theta 12/13 B. Csc theta= 12/13, sec theta=13/5 C. Csc theta= 13/12, sec theta=15/5 D. Csc theta= 5/12, sec theta=13/5
anonymous
  • anonymous
a = 9, b = 12, c = 15 and, If "theta" is the angle between a & b, then cos(theta) = 15/12 = 5/4 and, sin(theta) = 15/9 = 5/3 So, sec(theta) = 1/cos(theta) = 1/(5/4) = 4/5 csc(theta) = 1/sin(theta) = 1/(5/3) = 3/5
anonymous
  • anonymous
@Johnica Those are none of my answer choices

Looking for something else?

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

More answers

anonymous
  • anonymous
?
anonymous
  • anonymous
csc = 9/15 sec = 12/15
anonymous
  • anonymous
i am confused it still doesn't match my any of my answer choice is there something else i have to do?
anonymous
  • anonymous
?
anonymous
  • anonymous
ok is it D?
anonymous
  • anonymous
@Johnica
anonymous
  • anonymous
Are you there?
anonymous
  • anonymous
someone help me?@anyone
anonymous
  • anonymous
@izzy529
anonymous
  • anonymous
fine

Looking for something else?

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