Question

If sin Θ = 5 over 6 , what are the values of cos Θ and tan Θ?

Answer 1

cos would be 3.31 over 6 and tan would be 5 over 3.31

Answer 2

I don't necessarily want the answers; I would like to know how you get that thoo...

Answer 3

do want me to go step by step with ya?

Answer 4

because sin = opposite over hypotenuse, so thats the 5 over 6, that means that since c^2= a^2+b^2 that 6^2=5^2+b^2 so then 36=25+b^2 so b^2=11 and the square root of 11 is 3.31. so since the cos is adjacent over hypotenuse, that would be 3.31 over 6 and tan would be 5 over 3.31

Answer 5

I can show you how far I got??

Answer 6

sorry i was trying to type that as fast as i could

Answer 7

does that make sense?

Answer 8

Yes! That makes sense! :) Can you go step-by-step with me tho? I just want to make sure I'm doing it right.

Answer 9

yeah not a problem

So sin is equal to the opposite side over the hypotenuse, cos is equal to adjacent over hypotenuse and tangent is equal to opposite over adjacent.

since sin is 5 over 6, you can use the equation c^2 = a^2+b^2
c is the hypotenuse so 6
a can be either side but since we have 5 we will use 5, and we are trying to find b
so we have 6^2=5^2+b^2
that is 36=25+b^2
so subtract 25 from both sides and you get b^2=11
take the square root of both sides and you get b=3.31
b is now the other side we didn't have before (the adjacent side)
so now we have all the sides

Answer 10

Okay, I got that far. :)

Answer 11

So, now I just use SOH CAH TOA right?

Answer 12

yeah, just use SOH CAH TOA

Answer 13

Thats how I imagined it. I'm just stuck on how you knew where theta was? Is it because we were given sine then you backtracked from there?

Answer 14

I think I have the answers

Answer 15

yeah, just draw every triangle, you know that the hypotenuse is always the biggest side, so you have that and from there you can really use any angle, i just typically use that one and find the opposite side and the adjacent side

Answer 16

Okay, thanks.
Now, I just have one last question for this problem:

Answer 17

I got:

Answer 18

go ahead

Answer 19

\[\cos= \frac{ \sqrt{11} }{ 6 }

Tan= \frac{ 5 }{ \sqrt{11} }\]

Answer 20

yeah thats right

Answer 21

But these are my list choices:

Answer 22

cos Θ = ± 11 over 36 ; tan Θ = ± 1 over 11

cos Θ = ± 11 over 6 ; tan Θ = 11 over 6

cos Θ = ± square root 11 over 6 ; tan Θ = negative 1 over 11

cos Θ = ± square root 11 over 6 ; tan Θ = ± 5 square root 11 over 11

Answer 23

(Sorry, it took so long because my computer froze)

Answer 24

hmmm. im not sure where the 1 could come from... and the second one is not right because the 6 is the hypotenuse and doesn't belong in the tangent. so I'm thinking the last one, but i don't see how they could have 5 square root 11 over 11

Answer 25

That's what I'm saying! lol

Answer 26

it would be different if 11 had some perfect square factors so that you could simplify it, but thats not what they did, it really doesn't make sense because they gave you 5, so it shouldn't changed throughout the problem...?

Answer 27

Hmm, I'm not sure. I'll talk to my teacher about it in class tomorrow. but thank you so much for your help and patience!

Would you happen to have time to help with a few more problems??

Answer 28

Yeah not a problem go ahead

Answer 29

Okay, I'll open a new thread.

Answer 30

okay

Answer 31

just tag me in it when you make it