An acute angle θ is in a right triangle with sin θ = eight ninths. What is the value of cot θ?
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- anonymous

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- anonymous

Remember that cot=cos/sin.

- anonymous

|dw:1438145286805:dw|

- anonymous

Use Pythagorean Theorem to solve for the other side.

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## More answers

- anonymous

You there?

- anonymous

would it be sq 9^2-8^2

- anonymous

Yes.

- anonymous

thats none of the answer choices

- anonymous

Oh no, that's one of steps; not the answer. We're getting near, don't worry.

- anonymous

|dw:1438145770329:dw|

- anonymous

|dw:1438145807460:dw|

- anonymous

|dw:1438145868345:dw|

- anonymous

So what is the sine? What is the cosine?

- anonymous

\[\sin=\frac{ opposite }{ hypotenuse }; cosine=\frac{ adjacent }{ hypotenuse }\]

- anonymous

|dw:1438146075606:dw|

- anonymous

8/sq17?

- anonymous

What is the sine?

- anonymous

that is the sine

- anonymous

No. sq 17 is the adjacent, not the hypotenuse.You need opposite/hypotenuse.

- anonymous

|dw:1438146386165:dw|

- anonymous

oh then 8/9

- anonymous

Yes, and what is the cosine? Cosine is adjacent/hypotenuse.

- anonymous

Base you answer on the drawing. :)

- anonymous

sq17/9

- anonymous

Perfect! So now, we need to find cot.

- anonymous

cot=cos/sin.

- anonymous

|dw:1438146716415:dw|

- anonymous

Can you hover on the drawing I just drew, and click "Reply using drawing" and write the cos and sin?

- anonymous

What is the cos again? How about the sine?

- anonymous

|dw:1438146964481:dw|

- anonymous

Im confused because isnt the bottum line supposed to be opposite

- anonymous

and im so sorry my computer is being really dumb

- anonymous

Huh? Can draw the write the cos/sin? :)

- anonymous

cot=cos/sin

- anonymous

The bottom line of the triangle is supposed to be opposite, you put adjacent. and okay i will

- anonymous

cot=sq17/9 / 8/9

- anonymous

No. :(

- anonymous

okay, i give up lol

- anonymous

I was trolling. :(

- anonymous

XD

- anonymous

You're right! HAHA.

- anonymous

ohh! yay! lol

- anonymous

|dw:1438147400024:dw|

- anonymous

So now, to divide this, you multiply the numerator to the reciprocal of the denominator.

- anonymous

What is the reciprocal of 8/9?

- anonymous

|dw:1438147459855:dw|

- anonymous

i was wondering if the triangle should be like this tho

- anonymous

No. I put the angle in the|dw:1438147529285:dw|

- anonymous

Oh i see now

- anonymous

|dw:1438147560921:dw|

- anonymous

yeah i didnt notice you did that lol

- anonymous

So the reciprocal of 8/9 is 9/8.

- anonymous

yep

- anonymous

|dw:1438147658966:dw|

- anonymous

sq17/8!

- anonymous

Hooray!!! :D Great job!

- anonymous

Youre amazing!

- anonymous

I sent you message, and you didn't reply. :(

- anonymous

Sorry im doing an exam :( ill reply as soon as i am done! im struggling sorry.

- anonymous

Whoops! Sorry. :) Good luck!

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