# Thread: cot tan help °.

1. ## cot tan help °.

Hello! I'm new to the forum, and really really happy I found it!
I couldn't find an intro section, so thought I'd say hello here .

Now.
I understand if someone asks, what is the sin(36°)?
But if I have a problem
cot(68°36') what does that mean? Does it mean the cot of 68.36°? Or the supplementary of 68°, and the complimentary of 36°? And how would I solve for it? By taking the tan of the supplementary and complimentary?

Any help is greatly appreciated! Thank you!

-Nichole.

2. Originally Posted by Auri
Hello! I'm new to the forum, and really really happy I found it!
I couldn't find an intro section, so thought I'd say hello here .

Now.
I understand if someone asks, what is the sin(36°)?
But if I have a problem
cot(68°36') what does that mean? Does it mean the cot of 68.36°? Or the supplementary of 68°, and the complimentary of 36°? And how would I solve for it? By taking the tan of the supplementary and complimentary?

Any help is greatly appreciated! Thank you!

-Nichole.
This is a notation for degree's minutes and seconds

a minute (')is 1/60 of a degree and a second ('') is 1/60 of a minute.

So you have 68 degrees and 36 minutes this is the same as

$68+\frac{36}{60}=68.6^\circ$

3. Originally Posted by TheEmptySet
This is a notation for degree's minutes and seconds

a minute (')is 1/60 of a degree and a second ('') is 1/60 of a minute.

So you have 68 degrees and 36 minutes this is the same as

$68+\frac{36}{60}=68.6^\circ$
Oh! Now I understand! Thank you. And the cot of 68.6° would be the tan of the reference angle?

Edit - I got it now, it's 1/tan(68.6) !