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Mathematics 6 Online
OpenStudy (anonymous):

Help please. No idea about where to start! problem is in the first comment!

OpenStudy (anonymous):

OpenStudy (anonymous):

ok \[\cot ^{-1}x=\frac{ cosx }{ sinx }\]

OpenStudy (anonymous):

You are looking for: theta = arc cot(11.4/12)

OpenStudy (anonymous):

diameter = 22.8....so radius = (1/2) diameter = 11.4 height = 12

OpenStudy (anonymous):

Then the radius is half the diameter. So, \[\frac{ 22.8 }{ 2 }=r=11.4\]

OpenStudy (anonymous):

|dw:1384064221203:dw|

OpenStudy (anonymous):

you want theta...suggestion @Abdallah2012 ?

OpenStudy (anonymous):

Then just plug in 11.4 for r and 12 for h. so... \[\theta=\cot ^{-1}(\frac{ 11.4 }{ 12 })= about 46.47 ^{o}\]

OpenStudy (anonymous):

@Easyaspi314 If it is not much trouble I would appreciate it!

OpenStudy (anonymous):

@donny471 I also could have given an angle...I wanted @Abdallah2012 to get that angle.

OpenStudy (anonymous):

@Abdallah2012 Do you agree with that equation before we solve for it?

OpenStudy (anonymous):

@Easyaspi314 looks good to me

OpenStudy (anonymous):

You agree with my diagram?

OpenStudy (anonymous):

cot is adjacent leg/opposite leg

OpenStudy (anonymous):

so I put 11.4 next to theta, and 12 at the opposite leg

OpenStudy (anonymous):

agree with the diagram?

OpenStudy (anonymous):

@Easyaspi314 Yea

OpenStudy (anonymous):

@Abdallah2012 Do you agree with the diagram?

OpenStudy (anonymous):

so we want an angle whose cot is 11.4/12. But the easiest way...because there is no cot^-1 button on your calcualator. So we compute tan^-1 of (12/11.4) which is the same as computing cot ^-1 (11.4/12).

OpenStudy (anonymous):

so press tan^-1 button on your calculator and enter 12/11.4, you get approximately 46.47 degrees.

OpenStudy (anonymous):

Make sure your calculator is set in DEG mode.

OpenStudy (anonymous):

@Abdallah2012 Does this make 100% sense?

OpenStudy (anonymous):

@Easyaspi314 It actually does. you provided a very nice explanation on how to do it and it actually makes since to me. from looking at the same problem now it does not look half as bad as it did 5 minutes ago. Thank you very much! I truly appreciate it especially how you have me attempt to do it while you guide me through it instead of just providing an answer. So again thank you!

OpenStudy (anonymous):

Very welcome.

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