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Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.1hint: tan(B) = sin(B)/cos(B)

Albert0898
 one year ago
Best ResponseYou've already chosen the best response.1I did that already and am not getting any of the answers provided

Albert0898
 one year ago
Best ResponseYou've already chosen the best response.1If tan B = 9/13 and sin B = 9/13, then cos B = ?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0From @Vocaloid's hint, \[\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \qquad \implies \qquad \cos(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Just plug in your values and solve, simple

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos(\theta) = \frac{\sin(\theta)}{\tan(\theta)}\]

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.1actually I think there's a typo in the question, cos(B) = 1

Albert0898
 one year ago
Best ResponseYou've already chosen the best response.1Using that formula yields me an answers that is not provided

Albert0898
 one year ago
Best ResponseYou've already chosen the best response.1I've tried every formula today and I don't get any answer that resembles any option OpenStudy was my last resort lol

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.1you're right, it seems theres a typo in the question

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1weird question i guess we should use logic here LOL \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\] and we know hypotenuse is the longest side of right triangle cos = adjacent /hyp adjacent must be less than hypdw:1443408580125:dw

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1and voc is right there is a typo bec in the explanation tan=9/10 not 13 http://prntscr.com/8la7cy
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