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Ammarah
 4 years ago
how do u know when to use sin and cos?
Ammarah
 4 years ago
how do u know when to use sin and cos?

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IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0sin for opposite over hypotenus cos for adjacent over hypotenus

Ammarah
 4 years ago
Best ResponseYou've already chosen the best response.0I know but like in a problem...i dont get which one i should use

Ammarah
 4 years ago
Best ResponseYou've already chosen the best response.0Use a sine or cosine ration to find the value of each variable. Round decimals to the nearest tenth.

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0use that. sorry. i ahve to go soon.

Ammarah
 4 years ago
Best ResponseYou've already chosen the best response.0it doesnt help please solve these for me

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1358230939592:dw Refer to the reference angle. Sin => opposite side (to the reference angle) / hypotenuse In this diagram, can you determine what sine theta is?

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.1what is the opposite side to the reference angle theta?

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.1and what is the hypotenuse?

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.1Sorry : that was supposed to be z : If you know x and z, you can find theta. Because sin (theta) = x/z Got it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ammara, so in an equation, Callisto is saying is that in the figure above, \[\sin(\theta) = \frac{ x }{ z }\] So, in your figure, the first problem: you know the hypotenuse (18) and the angle opposite x (32 degrees). So you know that, \[\sin(32) = \frac{ x }{ 18 }\] Then solve for x :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And that's the sine of 32 degrees of course, not radians.

Ammarah
 4 years ago
Best ResponseYou've already chosen the best response.0what do i do for why then?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, you know that the angle opposite x is 32 degrees, and that the angle opposite the hypotentuse is 90 degrees. You also know the total sum of all interior angles of a triangle  so from that you can figure out the remaining angle you don't know, and do the same thing.

Ammarah
 4 years ago
Best ResponseYou've already chosen the best response.0i dont get it....how to i solve for y?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Or alternatively, use the cosine which is the adjacent angle over the hypotenuse: \[\cos(\theta) = \frac{ y }{ z }\] Substitute for the adjacent angle (32 degrees again) and for the hypotenuse (18), and solve for y the same way you solved for x above.

Ammarah
 4 years ago
Best ResponseYou've already chosen the best response.0Ok how about the next triangle...

Ammarah
 4 years ago
Best ResponseYou've already chosen the best response.0so in the left triangle were solving for a.

Ammarah
 4 years ago
Best ResponseYou've already chosen the best response.0how would u solve it? please solve it for me.......

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Correct. And you know that the cosine of the angle (48 degrees) is the same as 10 divided by a  careful here, because a is the hypotenuse: \[\cos(\Theta) = \frac{ 10 }{ a }\] So, \[\cos(48) = \frac{ 10 }{ a } \]

Ammarah
 4 years ago
Best ResponseYou've already chosen the best response.0how would i solve be now...i got 14.9 for a

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's correct. Now solve the last figure?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You got it. Can you do the last triangle?
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