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haileyy0401 Group Title

How do you figure out the csc and sec? For a standard-position angle determined by the point (x, y), what are the values of the trigonometric functions? For the point (9, 12), find csc theta and sec theta

  • one year ago
  • one year ago

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  1. AtiFS Group Title
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    well, If we draw these points on the graph, it would give us a right-angled triangle and then we may apply pythagoras' theorem to get "sin (theta)" & "cos (theta)" and "sec (theta) = 1/cos(theta)" & "csc(theta) = 1/sin(theta)". From the points (as it makes a right-angled triangle), we know Suppose, a = horizontal side of triangle, b = verticle, and c = longest side (or diagonal) So, a = 9, b = 12 and applying pythagoras' theorem now, (the theorem is c^2 = a^2 + b^2) we get c^2 = 9^2 + 12^2 c^2 = 81 + 144 c^2 = 225 c = 15. So, now the values of the sides of the triangle are 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 That's the answer. HTH :)

    • one year ago
  2. AtiFS Group Title
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    Is that the correct answer? can u verify?

    • one year ago
  3. haileyy0401 Group Title
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    I will be able to verify in one minute.

    • one year ago
  4. haileyy0401 Group Title
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    Thats not one of the choices.... a. csc = 15/12 sec = 15/9 b. csc = 15/12 sec = 12/15 c. csc = 9/15 sec = 12/15 d. csc = 9/12 sec = 9/15

    • one year ago
  5. haileyy0401 Group Title
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    oh did you mean csc and sec? instead of cos and sin?

    • one year ago
  6. AtiFS Group Title
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    yes

    • one year ago
  7. AtiFS Group Title
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    Option C is the correct answer

    • one year ago
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