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

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11 months ago
#1
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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#2
It might just be my student room, but it appears your message along with the title is not showing.
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10 months ago
#3
(Original post by 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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10 months ago
#4
(Original post by _Mia101)
shadowdweller the original poster has deleted their post. Could the tread be deleted? Thanks.
Looks like the thread just needed approving! Should be visible now
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10 months ago
#5
(Original post by 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.
Last edited by mqb2766; 10 months ago
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10 months ago
#6
(Original post by 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.
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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10 months ago
#7
(Original post by 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.)

These videos are great
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10 months ago
#8
(Original post by Sir Cumference)
These videos are great
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 months ago
#9
(Original post by 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...
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!
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10 months ago
#10
Looks like the thread just needed approving! Should be visible now
ah okay
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10 months ago
#11
(Original post by 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!
Not sure about that, but I was happy with them at the time. They do reflect my unusual preferences with respect to the AS, like a heavy and early emphasis on transformations done in a slightly non-standard way, but I'm leaving them up there for anyone who does want to make use of them.

My voice is a bit drearier than I remember, though.
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10 months ago
#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.

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10 months ago
#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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