Question

Find x.
A. 4
B. 4 times the square root of 2
C. 8
D. 4 times the square root of 3

Answer 1

SOH CAH TOA?

Answer 2

\[\large \sin(\theta) =\cfrac{opp}{hyp}\]\[\large \tan(\theta) = \cfrac{opp}{adj}\]\[\large \cos(\theta) = \cfrac{adj}{hyp}\]Which one can you apply here?

Answer 3

one with a hyp

Answer 4

maybe sin or cos

Answer 5

Good, that eliminates \(\tan (\theta)\)

Answer 6

Would it be \(\sin(45)=\cfrac{opp}{hyp}\) or \(\cos(45) = \cfrac{adj}{hyp}\)?

Answer 7

hmmmm i think cos

Answer 8

Good :)

Answer 9

what can we plug in for the \(\cfrac{adj}{hyp}\)?

Answer 10

x/18= 45

Answer 11

Correction: \[\large \cos(45) = \cfrac{x}{18}\] don't forget the trig part!

Answer 12

We can also rewrite it like \(\cfrac{\cos(45)}{1}= \cfrac{x}{18}\)

Answer 13

Now how would we solve for x?

Answer 14

cos(45)*18= x

Answer 15

right?

Answer 16

Great! you got this :D

Answer 17

9.45?

Answer 18

Well, not exactly..

Answer 19

Do you know what cos(45) =?

Answer 20

.707

Answer 21

Yes, and that is equivalent to \(\cfrac{\sqrt{2}}{2}\)

Answer 22

would i then equal one or srt2

Answer 23

Hmm.. one minute. I am not getting any of those answers after i simplified this.

Answer 24

I'm probably wrong I'm not good at simplifying

Answer 25

Any ideas on where you might have gone wrong? Don't feel like scrolling around through the whole thing checking.

Answer 26

We're looking for x, no?

Answer 27

ya

Answer 28

that was a ya to were looking for x not for the scrolling around part

Answer 29

What I was getting was that we found x to be = \(\cos(45) \cdot 18\) = \(\cfrac{\sqrt{2}}{2}\cdot 18\) = \(9 \sqrt{2}\) but that isnt one of your answers.

Answer 30

\[Cos(45°)= \frac{x}{18}\]\[x=18*\frac{\sqrt{2}}{2}=9*\sqrt{2}\]

Answer 31

That is correct. The question's wrong.

Answer 32

Well the given choices might be wrong

Answer 33

Yeah... That's what I was thinking.

Answer 34

ya it is right thanks you guys

Answer 35

no problem