# Sin, cos and tan of an angle above 90°

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This is what I believe to be true (bearing in mind I have only recently started studying trigonometry for GCSE):

A right triangle is made up of 3 angles. A right angle and 2 acute angles.

The sin of an angle is the ratio between a right triangles opposite side and its hypotenuse.

The cos of an angle is the ratio between a right triangles adjacent side and its hypotenuse.

The tan of an angle is the ratio between a right triangles opposite side and its adjacent side.

Now here is my question:

How can you find the sin, cos or tan of an angle at 90° or more?

As I said before a right triangle is made up of 1 right angle and 2 acute angles. That means that the other 2 angles cannot be 90° or more.

So I am going to ask my question again just to clarify what my question is. How can you find the sin, cos or tan of an angle in a right triangle if it is 90° or more (seen as these angles in a triangle cannot possibly exist)?

I hope I have made my question clear, please answer my question in as simple terms possible seen as (as I previously said) I have only recently started doing trigonometry.

Thank you.

A right triangle is made up of 3 angles. A right angle and 2 acute angles.

The sin of an angle is the ratio between a right triangles opposite side and its hypotenuse.

The cos of an angle is the ratio between a right triangles adjacent side and its hypotenuse.

The tan of an angle is the ratio between a right triangles opposite side and its adjacent side.

Now here is my question:

How can you find the sin, cos or tan of an angle at 90° or more?

As I said before a right triangle is made up of 1 right angle and 2 acute angles. That means that the other 2 angles cannot be 90° or more.

So I am going to ask my question again just to clarify what my question is. How can you find the sin, cos or tan of an angle in a right triangle if it is 90° or more (seen as these angles in a triangle cannot possibly exist)?

I hope I have made my question clear, please answer my question in as simple terms possible seen as (as I previously said) I have only recently started doing trigonometry.

Thank you.

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It might just be my student room, but it appears your message along with the title is not showing.

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#3

(Original post by

It might just be my student room, but it appears your message along with the title is not showing.

**Mrepic Foulger**)It might just be my student room, but it appears your message along with the title is not showing.

shadowdweller the original poster has deleted their post. Could the tread be deleted? Thanks.

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#4

(Original post by

shadowdweller the original poster has deleted their post. Could the tread be deleted? Thanks.

**_Mia101**)shadowdweller the original poster has deleted their post. Could the tread be deleted? Thanks.

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#5

(Original post by

This is what I believe to be true (bearing in mind I have only recently started studying trigonometry for GCSE):

A right triangle is made up of 3 angles. A right angle and 2 acute angles.

The sin of an angle is the ratio between a right triangles opposite side and its hypotenuse.

The cos of an angle is the ratio between a right triangles adjacent side and its hypotenuse.

The tan of an angle is the ratio between a right triangles opposite side and its adjacent side.

Now here is my question:

How can you find the sin, cos or tan of an angle at 90° or more?

As I said before a right triangle is made up of 1 right angle and 2 acute angles. That means that the other 2 angles cannot be 90° or more.

So I am going to ask my question again just to clarify what my question is. How can you find the sin, cos or tan of an angle in a right triangle if it is 90° or more (seen as these angles in a triangle cannot possibly exist)?

I hope I have made my question clear, please answer my question in as simple terms possible seen as (as I previously said) I have only recently started doing trigonometry.

Thank you.

**finnan04**)This is what I believe to be true (bearing in mind I have only recently started studying trigonometry for GCSE):

A right triangle is made up of 3 angles. A right angle and 2 acute angles.

The sin of an angle is the ratio between a right triangles opposite side and its hypotenuse.

The cos of an angle is the ratio between a right triangles adjacent side and its hypotenuse.

The tan of an angle is the ratio between a right triangles opposite side and its adjacent side.

Now here is my question:

How can you find the sin, cos or tan of an angle at 90° or more?

As I said before a right triangle is made up of 1 right angle and 2 acute angles. That means that the other 2 angles cannot be 90° or more.

So I am going to ask my question again just to clarify what my question is. How can you find the sin, cos or tan of an angle in a right triangle if it is 90° or more (seen as these angles in a triangle cannot possibly exist)?

I hope I have made my question clear, please answer my question in as simple terms possible seen as (as I previously said) I have only recently started doing trigonometry.

Thank you.

trig obtuse angles

And pick one that is readable/watchable. For instance

https://amsi.org.au/teacher_modules/...gonometry.html

Two basic ways of imagining it is

* apply the sin and cos rules for non right angled triangles.

* think about the complementary right angled triangle in the second quadrant.

Have a read/think first, then ask about any problems.

Last edited by mqb2766; 10 months ago

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#6

**finnan04**)

This is what I believe to be true (bearing in mind I have only recently started studying trigonometry for GCSE):

A right triangle is made up of 3 angles. A right angle and 2 acute angles.

The sin of an angle is the ratio between a right triangles opposite side and its hypotenuse.

The cos of an angle is the ratio between a right triangles adjacent side and its hypotenuse.

The tan of an angle is the ratio between a right triangles opposite side and its adjacent side.

Now here is my question:

How can you find the sin, cos or tan of an angle at 90° or more?

As I said before a right triangle is made up of 1 right angle and 2 acute angles. That means that the other 2 angles cannot be 90° or more.

So I am going to ask my question again just to clarify what my question is. How can you find the sin, cos or tan of an angle in a right triangle if it is 90° or more (seen as these angles in a triangle cannot possibly exist)?

I hope I have made my question clear, please answer my question in as simple terms possible seen as (as I previously said) I have only recently started doing trigonometry.

Thank you.

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#7

(Original post by

Here are a couple of videos I made for my students about exactly this issue. (I make no claims that these videos are any good - they're certainly not slick or flashy! - but they say what I wanted to say.)

**Pangol**)Here are a couple of videos I made for my students about exactly this issue. (I make no claims that these videos are any good - they're certainly not slick or flashy! - but they say what I wanted to say.)

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#8

(Original post by

These videos are great

**Sir Cumference**)These videos are great

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#9

(Original post by

You are very kind. I made these after the first year of the new A Level spec, where I had done a bad job of timing the course and we didn't finish the AS in time. I thought that for the second attempt at the AS, I'd make videos like this for the whole course, ask students to watch them before the lessons, and hopefully speed everything up. Didn't work, too many of them didn't watch them, thus the project was abandoned part way through the AS pure content...

**Pangol**)You are very kind. I made these after the first year of the new A Level spec, where I had done a bad job of timing the course and we didn't finish the AS in time. I thought that for the second attempt at the AS, I'd make videos like this for the whole course, ask students to watch them before the lessons, and hopefully speed everything up. Didn't work, too many of them didn't watch them, thus the project was abandoned part way through the AS pure content...

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#10

(Original post by

Looks like the thread just needed approving! Should be visible now

**shadowdweller**)Looks like the thread just needed approving! Should be visible now

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#11

(Original post by

Ah that's a shame. If the rest of your topic videos were as good as those then I'm sure your channel would have taken off and Examsolutions would have had a new rival!

**Sir Cumference**)Ah that's a shame. If the rest of your topic videos were as good as those then I'm sure your channel would have taken off and Examsolutions would have had a new rival!

My voice is a bit drearier than I remember, though.

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#12

That is a really lovely question, and you're right, if that's how your GCSE materials define them, no wonder you would ask this. The more general (and natural) definition of trig functions comes from thinking about how a point on the edge of a unit circle rotates. "Unit" here means it has radius of length 1, and links to the fact that as you rightly say, trig functions are to do with proportions of sides of a triangle. What is the link between circles and triangles?

Imagine the circle is centred at (0,0), and then your x and y coordinates of a random point on the circle give you 2 lengths, or sides. You can also draw a line from the origin to the point on the circle. The shape of these 3 lines will always be a right triangle as the x and y coordinates are always perpendicular. If you are confused right now I don't blame you as my description was quite poor but this video will do a much better job.

https://www.youtube.com/watch?v=ZffZvSH285c

Imagine the circle is centred at (0,0), and then your x and y coordinates of a random point on the circle give you 2 lengths, or sides. You can also draw a line from the origin to the point on the circle. The shape of these 3 lines will always be a right triangle as the x and y coordinates are always perpendicular. If you are confused right now I don't blame you as my description was quite poor but this video will do a much better job.

https://www.youtube.com/watch?v=ZffZvSH285c

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#13

Oh sorry I didn't refresh my page before this was answered. Making ur own vids>linking khan academy lol

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