Question

I need help with #5 & #6. My attempts will be in comments.

Answer 1

!

Answer 2

you have tangent correct so far... the cotangent is 1/tan or adjacent/opposite

Answer 3

the definitions are here on the first page

Answer 4

would it be 7/5? @DanJS

Answer 5

yes, if it were for that same angle x as before... it says Cot(y) this time

Answer 6

@DanJS Awesome!! Now, Could you guide me through 6?

Answer 7

i see whatever cos is i flip it to make it sec

Answer 8

since it is a right triangle, angles x+y=90
notice y=90-x
and

\[\cot(90-x) = \tan(x)\]

Answer 9

whoa, okay.

Answer 10

listed under cofunctions on that pdf
cot(y) = cot(90-x) = tan(x)

Answer 11

sin and cos involve the hypotenuse, you should figure that length out first

Answer 12

nm, you have it from another prob you did

yes, flip cos over, sec(x)=1 / cos(x)

however, the secant is for y, not x

Answer 13

i got

Answer 14

\[\sec(y) = \frac{ 1 }{ \cos(y) }=\frac{ 1 }{ 5/\sqrt{74} }=\frac{ \sqrt{74} }{ 5 }\]

Answer 15

cos(x) is just adj/hp = 7/root(74)

Answer 16

Yay! i was about to put does cosx=7/square root 74.

Answer 17

Now, if you have a second.. did you look over #6 did i do ok on that?

Answer 18

yep, and cos(y) = 5/root(74), so sec(y) is that flipped over

Answer 19

Sweet!

Answer 20

6 looks blank , but that is what i just did, cos(x) and sec(y)

Answer 21

omg! lol i meant 4! @DanJS

Answer 22

yes, looks good

Answer 23

I appreciate it!! You rock!!

Answer 24

welcome, gl