All possible values of theta

[Relz]

Question: "θ is an angle in standard position with domain 0° <θ< 360° and cot θ = 24/-7. Find all possible values of θ"

Work: I have found the hypotenuse to be 25, through pythagoreans theorem. I know that there are only two possible values of theta: one in quatrant II and one in quadrant IV from the CAST rule. I'm not really sure how to find the angles though. I tried flipping the ratio and using the tangent button on my calculator but they come out the same and at -0.005 degrees which I know is way off! I know that i've learned this before but I can't seem to remember. Can anyone help?

 

[renegade05]

Question: "θ is an angle in standard position with domain 0° <θ< 360° and cot θ =

. Find all possible values of θ"

Work: I have found the hypotenuse to be 25, through pythagoreans theorem. I know that there are only two possible values of theta: one in quatrant II and one in quadrant IV from the CAST rule. I'm not really sure how to find the angles though. I tried flipping the ratio and using the tangent button on my calculator but they come out the same and at -0.005 degrees which I know is way off! I know that i've learned this before but I can't seem to remember. Can anyone help?

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[Relz]

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Sorry, that wasnt supposed to be a link. I fixed it so the answer is shown. Any ideas?

 

[mmm4444bot]

You should not be taking the tangent of 7/24 (to find the reference angle) because 7/24 is not an angle.

7/24 is a trig ratio; you want to find the angle whose tangent is 7/24.

Use the inverse-tangent function (arctangent).

 

[Relz]

You should not be taking the tangent of 7/24 (to find the reference angle) because 7/24 is not an angle.

7/24 is a trig ratio; you want to find the angle whose tangent is 7/24.

Use the inverse-tangent function (arctangent).

Yes I realize that I need the angle and that these are the sides. Thats why I found the hypotenuse of it. I dont have arctan on my calculator and my calculator also doesnt realize the CAST rule, obviously. So I cant just plug it in and get the answer. Is there a way I can find the first angle then subtract it from 180, then find the second angle?

 

[mmm4444bot]

I realize that I need the angle and that these are the sides

Okay, but you misinterpreted the tangent function.

I tried flipping the ratio and using the tangent button on my calculator but they come out the same and at -0.005 degrees

The tangent function does not output angles.

Hence, tan(-7/24) = 0.005 degrees is not the correct interpretation.

This actually goes with a reference triangle having opposite side 1 and adjacent side 200 and the angle is 7/24ths of one degree.

Not what you want.

---

You want:

arctan(-7/24) = -16.2567 degrees

because it is the inverse trig function "arctangent" that accepts some angle's tangent value as input and then outputs the angle itself.

I dont have arctan on my calculator

Are you certain? What model calculator is this?

Windows has a scientific calculator built-in. Alternatively, you can google for inverse-function calculators on-line, like this one.

 

[Relz]

Okay, but you misinterpreted the tangent function.





The tangent function does not output angles.

Hence, tan(-7/24) = 0.005 degrees is not the correct interpretation.

This actually goes with a reference triangle having opposite side 1 and adjacent side 200 and the angle is 7/24ths of one degree.

Not what you want.

---

You want:

arctan(-7/24) = -16.2567 degrees

because it is the inverse trig function "arctangent" that accepts some angle's tangent value as input and then outputs the angle itself.




Are you certain? What model calculator is this?

Windows has a scientific calculator built-in. Alternatively, you can google for inverse-function calculators on-line, like this one.

Well, even if I did have it, my teacher wouldn't accept arctan as we haven't learned it yet. I know it sounds silly and I agree but I know from experience that if I do something that we haven't learned yet, it's marked wrong. I guess we are supposed to learn the basics first and then use short cuts.

 

[Mrspi]

Well, even if I did have it, my teacher wouldn't accept arctan as we haven't learned it yet. I know it sounds silly and I agree but I know from experience that if I do something that we haven't learned yet, it's marked wrong. I guess we are supposed to learn the basics first and then use short cuts.

"arctan" means "the angle whose tangent is...."

We can say tan 60^o = (sqrt 3)/2

We can say "the angle whose tangent is (sqrt 3)/2 is 60^o"

Or, we can say "tan^-1 (sqrt 3)/2) = 60^o"

Or, we can say "arctan (sqrt 3)/2 = 60^o"

Now, I realize I've left out of this discussion things like "principal values" and "reference angles" ....but my purpose is to illustrate that you have most likely dealt with the concept of "an angle which has a certain value for a tangent."

And on your calculator, you have probably used the "second function key" to find an angle which has a certain value for a trig ratio.

Back in the day when there WERE no calculators, students used trig tables to find the values for things like sin 80^o, or find the angle whose cosine = 1/2......

I can't believe your teacher would object to using your calculator to find the angle once you have the value of one of the functions. You may well need to know about the signs of the trig functions in various quadrants, and about the periodicity of the trig functions as well.

 

[mmm4444bot]

my teacher wouldn't accept [using] arctan as we haven't learned it yet.

The inverse trig functions are not "short cuts".

Clearly, your teacher must have another method in mind.

Please ask your teacher how you're supposed to find the two values of theta (I think they are 163.74 degrees and 343.72 degrees) without using any inverse trig function, and then let me know. I would like to learn, too! :cool: