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anonymous
 one year ago
Assuming cos t = .45 and cos w = .89, both t and w are positive, and both t and w determine a terminal point in quadrant 1, then which of the following statements best describes the relationship between t and w?
A. t > w
B. w > t
C. It is not possible to tell from the given information.
anonymous
 one year ago
Assuming cos t = .45 and cos w = .89, both t and w are positive, and both t and w determine a terminal point in quadrant 1, then which of the following statements best describes the relationship between t and w? A. t > w B. w > t C. It is not possible to tell from the given information.

This Question is Closed

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Have you looked at the first quadrant and notice what happens as the cos(t) values increase? like what happens to t?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you are given t and w are positive

freckles
 one year ago
Best ResponseYou've already chosen the best response.1but that doesn't answer your question

freckles
 one year ago
Best ResponseYou've already chosen the best response.1https://www.mathsisfun.com/geometry/images/circleunit304560.gif Look at this picture what is cos(60) and cos(30) equal to?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos(60) = 1/2 and cos(30) = \[\sqrt{3}/2\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[60^o >30^o \\ \cos(60^o)=\frac{1}{2} ? \frac{\sqrt{3}}{2}=\cos(30^o)\] which of those values are greater ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1is 1 bigger than sqrt(3) or is sqrt(3) bigger than 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sqrt(3) is bigger than 1

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[60^o>30^o \\ \cos(60^o)<\cos(30^o) \\ \text{ and you have } \cos(t)<\cos(w)\] and since we know t and w are in the first quadrant just like 60deg and 30deg was then we know what about t and w?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1why? didn't we have that cos(60)<cos(30) but 60>30 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Was it because of the cosine?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The cosine of 60 is giving us a value that is less than our cosine of 30 I think

freckles
 one year ago
Best ResponseYou've already chosen the best response.1yes 1/2 is less than sqrt(3)/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the answer would be w > t (B)?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok see if you follow this: \[60^o>30^o \\ \text{ so } \cos(60^o)=\frac{1}{2}<\frac{\sqrt{3}}{2}=\cos(30^o) \\ \text{ we have} \cos(60^o)<\cos(30^o) \text{ but } 60^o>30^o \\ \text{ You are given } \cos(t)<\cos(w) \text{ so you can draw what conclusion about } t?w\] hint replace 60 with t and 30 with w

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you can also notice in the first quadrant as the angle increases the cos value of those angles decrease

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So if the equations decreased that would make t > w?
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