Question

The legs of a right triangle measure 5 inches and 7 units. if \(\theta\) is the angle between the 5-inch leg and the hypotenuse, cos \(\theta\) = ___.

Answer 1

@dude

Answer 2

Pythagorean Theorem to find hypotenuse
\(a^2+b^2=c^2\)
Do you want assistance or the answer? ( ͡° ͜ʖ ͡°)

Answer 3

Options: 0.07, 0.81, 0.58, and 0.71

Answer 4

The answer le ( ͡° ͜ʖ ͡°)

Answer 5

That goes against the rules though ( ͡° ͜ʖ ͡°)

Answer 6

Curses (ಥ ͜ʖಥ)

Answer 7

\(5^2+7^2=c^2\)

Answer 8

That would be... 74?

Answer 9

did I take a wrong turn at Albuquerque?

Answer 10

Alberqueque?

Answer 11

Have you seriously never watched bugs bunny?
https://www.youtube.com/watch?v=e8TUwHTfOOU

Answer 12

\(5^2+7^2=c^2\\
25+49=c^2\\
74=c^2\)
root on both sides
\(\large{\pm\sqrt{74}=c}\)

Answer 13

Hey, lets keep this on math ( ͡° ͜ʖ ͡°)

Answer 14

Right ( ͡° ͜ʖ ͡°)

Answer 15

Now what? o-o

Answer 16

The 5 and 7 could be on either side

Answer 17

mfw the 7 is shorter

Answer 18

Disclaimer: This is not drawn to scale

Answer 19

LOL

Answer 20

We can use cosine
\(cos(x)=\frac{Adjacent}{Hypotenuse}\)

Answer 21

*head explodes*

Answer 22

\(cos(\theta)=\frac{5}{\sqrt{74}}\)

Answer 23

Ah ok, forgot what adjacent meant there

Answer 24

That is nice (ಥ ͜ʖಥ)

Answer 25

So 0.58 is my answer?

Answer 26

How did you get that?

Answer 27

\(5\div\sqrt{74}\)
\(\sqrt{74}=8.60\)
\(5\div8.60=0.58\)

Answer 28

You have
\(cos(\theta)=0.58\)
Theta is still not isolated you have to do the opposite of cos
Dont know how to explain this x.x

Answer 29

@563blackghost I see you viewing the post, can you explain it? O-o

Answer 30

dude is doing pretty well in my opinion.

Answer 31

Do you know your ratios? for sin, tan, cos?

Answer 32

Not really no x.x

Answer 33

ಠ_ಠ

Answer 34

You need to know your ratios before actually completeing this problem.

Answer 35

How about the format of Pythagorean Theorem?

Answer 36

Do you mean \(a^2+b^2=c^2\)?

Answer 37

yesh dat :3

Answer 38

Do you know what theta is? ಠ_ಠ

Answer 39

How did you guess I didn't? ಠ_ಠ

Answer 40

Continue ghost >.<

Answer 41

okay...

Answer 42

Sorry I wrote the last part wrong....

So let's draw our right triangle....



There are three ratios for a right triangle....

\(\large\bf{\sin (x)=\frac{opposite}{hypotenuse}}\)

\(\large\bf{\tan (x)=\frac{adjacent}{hypotenuse}}\)

\(\large\bf{\cos (x)=\frac{adjacent}{hypotenuse}}\)

Answer 43

Each ratio is plugged according to a triangles angle or sides.



For example if we want to find the angle A we would use the sides around.

We see we have a side x and a side y.

`The hypotenuse will always be the hypotenuse`.

So `side y` is the hypotenuse.

Since `side x` is next to `angle A` it is our adjacent.

Which ratio contains a hypotenuse and adjacent?

Answer 44

sorry `side y` is meant to be typed as `side z`.

Answer 45

For example if we want to find the angle A we would use the sides around.

We see we have a `side x` and a `side z`.

The hypotenuse will always be the hypotenuse.

So `side z` is the hypotenuse.

Since `side x` is next to` angle A` it is our adjacent.

Which ratio contains a hypotenuse and adjacent?

Answer 46

corrected paragraph above ^

Answer 47

\(\Large\frac{x+y}{z}\)?

Answer 48

Close.

`side y` is not included due to only using `2 sides` for a triangle ratio.

So let's remove `side y`.

\(\Large\bf{\frac{x}{z}}\)

You formatted this correctly, which ratio is this identified as?

`Sin, Tan, or Cos`?

Answer 49

Judging by what you posted before, tan or cos?

Answer 50

(although I'm assuming cos)

Answer 51

Correct! Cosine!

Answer 52

So we would follow as:

\(\Large\bf{cos(\theta)=\frac{x}{z}}\)

We would invert this so we can find the angle.

\(\Large\bf{cos^{-1}(\frac{x}{z})=\theta}\)

Do you understand the process to finding an angle?

Answer 53

Not really no

Answer 54

What is the -1 for in that equation? (and how is that calculated?)

Answer 55

The `power of -1` is to invert the equation. That is all.

This way you can plug it into your calculator.

Answer 56

\[2^{-1}=\frac{1}{2}\]

Answer 57

invert the equation...?

Answer 58

How does one turn a math equation inside out and dissect it? x'D

Answer 59

by taking it to the power of -1.

Answer 60

\[2^{-2}=\frac{1}{4}\]

Answer 61

For example the inversion of 2 is `1/2` if we multiply the two to each other we would get `1`.

Answer 62

*brain explodes*

Does not compute... How does 2 = 1/2
what does inverting mean?

Answer 63

How do you calculate negative powers? O_o

Answer 64

hmmmm

Inversion means `to turn the opposite`.

\(\large\bf{a^{-n}=\frac{1}{n}}\)

This is the format of a number carried by a negative.

Answer 65

\[\frac{2}{1}=>\frac12\]
or
\(2^{-1}=\frac{1}{2}\)

Answer 66

Define opposite of a number? I can understand a... OH NOW THAT IT'S VISUALIZED I GET IT

Answer 67

We invert the equation we got to find what `theta` is.

\(\Large\bf{cos(\theta)=\frac{x}{z} \rightarrow cos^{-1}(\frac{x}{z})=\theta}\)

Answer 68

Let's take your question now.

We get that the legs are `5` and `7`.



We want to know the angle between `5` and the `hypotenuse`.



If we look at the triangle we see that `5 is the adjacent` and that `7 is the opposite`. Which ratio do we use?

Answer 69

Didn't we already determine it would be cosine?

(sorry for disappearing, dinner was done)

Answer 70

yes we did, but im trying to see if you understand after the ratios have been stated and an explanation of the use of it.

Answer 71

Does cosine equal what goes in the function? Or is there something else that goes on?

If thats the case, didn't you define tan as the same as cos? Doesn't that mean you would end with the same exact result?


PS: Sorry I'm so stupid x'D

Answer 72

There are two ways you can solve your problem.

You can solve by tan and/or cosine.

@Dude solved by finding the hypotenuse and then used the length of the hypotenuse and the adjacent to find the angle.

You can also use tangent by using the adjacent and the opposite.

Either way will end up with the same solution.

Answer 73

So if I'm understanding this right, at this point, it's this?
\(\Large^{-1}\frac{7}{5}\) or \(\Large\frac{5}{7}\) = \(\large5\div7\)
Then from there I get 0.71... is that correct?

Answer 74

@563blackghost ?

Answer 75

>-< @dude ?

Answer 76

Welp I read this all wrong
5 and 7 are different units. I feel dumb

Answer 77

@velcolied

Answer 78

@vocaloid *