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How do you figure out the csc and sec? For a standardposition 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
How do you figure out the csc and sec? For a standardposition 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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AtiFSBest ResponseYou've already chosen the best response.1
well, If we draw these points on the graph, it would give us a rightangled 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 rightangled 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

AtiFSBest ResponseYou've already chosen the best response.1
Is that the correct answer? can u verify?
 one year ago

haileyy0401Best ResponseYou've already chosen the best response.0
I will be able to verify in one minute.
 one year ago

haileyy0401Best ResponseYou've already chosen the best response.0
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

haileyy0401Best ResponseYou've already chosen the best response.0
oh did you mean csc and sec? instead of cos and sin?
 one year ago

AtiFSBest ResponseYou've already chosen the best response.1
Option C is the correct answer
 one year ago
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