Why can't we apply trig ratios to any angle of a right triangle?

In summary, trigonometric ratios are only defined for non-right angles of a right triangle. This is because the cosine of 90 degrees is zero and the tangent of 90 degrees is undefined. These ratios cannot be extended to any angle of a right triangle because it would result in an undefined value. While trigonometric functions can be defined for all values, they are based on the ratios defined for a right triangle. The concept of infinity is not applicable in this context, and attempting to use the ratios at 90 degrees would result in a shape that is not a triangle.
  • #1
musicgold
304
19
Why do trigonometric ratios have to be related to the angle between the base and hypotenuse of a right angle triangle?

I am trying to understand why I can't use these ratios to any angle of a right angle triangle. I try to do that in the attached document. It seems to work for all ratios except cos90 and tan90.

I know the origin of the trigonometric ratios is in the unit circle and the length of the cord, but why can't we extend this principal to any angle of a right angle triangle?

Thanks.
 

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  • #2
The cosine of 90 degrees is zero and the tangent of 90 degrees is infinite. What's wrong with that?
 
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  • #3
It is true that sin B = 1, because B = 90 degrees.

But I don't understand why you said "sin B = k/k".
 
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  • #4
The answer is that the trigonometric ratios are defined only for the non-right angles of a right triangle.
 
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  • #5
SteamKing said:
The cosine of 90 degrees is zero and the tangent of 90 degrees is infinite. What's wrong with that?

Cos B = adjacent side / hypotenuse

To find Cos B, I am not sure which adjacent side to use - it can be h or b.

Either way, the ratios ( h/k or b/k) are not going to be zero. So I was not sure what to do.


AlephZero said:
It is true that sin B = 1, because B = 90 degrees.

But I don't understand why you said "sin B = k/k".

Sin B = opposite side / hypotenuse

In this case both of them are k.

Chestermiller said:
The answer is that the trigonometric ratios are defined only for the non-right angles of a right triangle.

Ok. I didn't know that. Where can I read more on this?

Thanks.
 
  • #6
musicgold said:
Cos B = adjacent side / hypotenuse

To find Cos B, I am not sure which adjacent side to use - it can be h or b.

Either way, the ratios ( h/k or b/k) are not going to be zero. So I was not sure what to do.




Sin B = opposite side / hypotenuse

In this case both of them are k.



Ok. I didn't know that. Where can I read more on this?

Thanks.

Adjacent side = side adjacent to the angle = side 'coming out of the angle'.
Opposite side = side opposite the angle = side the angle does not 'touch'.

Isn't this material in your textbook or course notes? If not, see http://www.purplemath.com/modules/basirati.htm . Look at the diagram.
 
  • #7
musicgold said:
Cos B = adjacent side / hypotenuse

To find Cos B, I am not sure which adjacent side to use - it can be h or b.

Either way, the ratios ( h/k or b/k) are not going to be zero. So I was not sure what to do.




Sin B = opposite side / hypotenuse

In this case both of them are k.



Ok. I didn't know that. Where can I read more on this?

Thanks.
It should be in any plane geometry textbook. Maybe they never thought to mention that these definitions don't apply to the right angle of the triangle.

Chet
 
  • #8
Chestermiller said:
The answer is that the trigonometric ratios are defined only for the non-right angles of a right triangle.
That is true only for the "trigonometric ratios" as defined in a trig course, in terms of a right triangle. The trig functions, sin(x) and cos(x), are defined for all x. The tangent and cosec functions are defined for all x such that [itex]sin(x)\ne 0[/itex]. The cotangent and secant functions are defined for all x such that [itex]cos(x)\ne 0[/itex].

They can be defined completely independently of "triangles" using power series for example.
 
  • #9
HallsofIvy said:
That is true only for the "trigonometric ratios" as defined in a trig course, in terms of a right triangle. The trig functions, sin(x) and cos(x), are defined for all x. The tangent and cosec functions are defined for all x such that [itex]sin(x)\ne 0[/itex]. The cotangent and secant functions are defined for all x such that [itex]cos(x)\ne 0[/itex].

They can be defined completely independently of "triangles" using power series for example.
This, of course, is correct, but it doesn't seem to address the question in the context that OP was asking it.

Chet
 
  • #10
SteamKing said:
... tangent of 90 degrees is infinite.

This is not true. Tangent of 90 degrees is undefined.
 
  • #11
gopher_p said:
This is not true. Tangent of 90 degrees is undefined.

But it's still infinite.
 
  • #12
SteamKing said:
But it's still infinite.

Tangent of 90 degrees is not defined. It is not a thing. It is mathematically meaningless; like "length of a circle" or "square root of cow". It does not have properties. In particular it does not have the property of being infinite. I don't know who told you that it did, but they were wrong.
 
  • #13
When I look at a reference like Abramowitz and Stegun for the tangent of 90 degrees, I don't see 'undefined' or the 'square root of a cow' listed; I see a lazy 8, which to me means infinity. Now, I agree that mathematical operations involving infinity are undefined, but if the limit of a series tends to infinity, we don't say that the series is 'undefined', we say it 'diverges'.

BTW, I take the 'length of a circle' to means its circumference.
 
  • #14
Here's a good way to think about it: when you apply the ratios to right angles, essentially you're going to have a right triangle where the angle you have the ratio on is also a right angle. Now obviously this shape wouldn't actually be a triangle, but using the ratios at 90 degrees is more like seeing what the ratios approach when one of the two other angles approaches 90.
 
  • #15
SteamKing said:
When I look at a reference like Abramowitz and Stegun for the tangent of 90 degrees, I don't see 'undefined' or the 'square root of a cow' listed; I see a lazy 8, which to me means infinity.

This source also says ##\ln 0=-\infty##, ##e^\infty=\infty##, and ##e^{-\infty}=0##. It also defines the "circular functions" in terms of complex exponentials. It uses the lemniscate repeatedly, and yet I can't for the life of me find where they actually say what they mean when they use it. I'm going to go out on a limb and say that it's not entirely appropriate as a reference for precalculus mathematics.

Under just about any definition of tangent appropriate for non math majors, "tangent of ninety degrees" is virtually meaningless in that it does not have the same meaning as "tangent of x" for any other (appropriate) value of x.

Now, I agree that mathematical operations involving infinity are undefined, but if the limit of a series tends to infinity, we don't say that the series is 'undefined', we say it 'diverges'.

Right. We aren't talking about limits though. We're talking about right triangle trigonometry. If you want to talk about limits of real-valued functions, then you should say that. But since you brought up series, I'll remind you of the recent internet kerfuffle regarding ##\sum n=-1/12##. While there is a context in which that is true, it is not part of the "standard" interpretation of infinite sums. Furthermore, it would be misleading and (if you actually know what you're talking about) willfully dishonest to tell someone that the result of adding all of the positive integers together gives you -1/12.

What I am trying to convey to you is that it is dishonest and misleading to tell a precalculus student that tangent of 90 degrees is infinite.

BTW, I take the 'length of a circle' to means its circumference.

Oops. I meant to say sphere.
 

Related to Why can't we apply trig ratios to any angle of a right triangle?

1. Why can't we apply trig ratios to any angle of a right triangle?

Trigonometric ratios, such as sine, cosine, and tangent, are based on the relationship between the sides of a right triangle and its angles. These ratios only apply to right triangles because they are defined by the ratios of the sides, which are specific to right triangles. This means that these ratios cannot be applied to other types of triangles or shapes.

2. Can trig ratios be used for obtuse or acute angles in a right triangle?

No, trigonometric ratios can only be applied to right triangles, which have a 90-degree angle. Acute and obtuse angles do not have a defined relationship between the sides, so trig ratios cannot be used to find missing sides or angles in these types of triangles.

3. Why are trig ratios only applicable to right triangles?

Trigonometric ratios were developed specifically for right triangles because of their unique relationship between the sides and angles. The ratios were created to solve problems related to right triangles, such as finding missing sides or angles, and they are not applicable to other types of triangles.

4. Is there any exception where trig ratios can be used for non-right triangles?

No, there are no exceptions where trig ratios can be used for non-right triangles. These ratios are only valid for right triangles and cannot be applied to other types of triangles or shapes.

5. Can trig ratios be used for angles greater than 90 degrees in a right triangle?

No, trigonometric ratios only work for angles up to 90 degrees in a right triangle. This is because the sides and angles of a right triangle have a specific relationship that is defined by these ratios. Angles greater than 90 degrees do not follow this relationship, so trig ratios cannot be used to solve for them.

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