# point on circle below a given point

This topic is 3332 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

I've spent some time and made a detailed picture / diagram of what I need to solve http://s661.photobucket.com/albums/uu335/zaneski13/?action=view¤t=mathproblemcam.jpg So look at that picture and if you can find a methood to get the angle of the triangle thats below the top triangle ( read note 2 on the image ) then please share Once I know that angle I can do everything else Thanks

##### Share on other sites
Your target's X coordinate would be
xt = r.tan(θ)
and so its y coordinate would be
yt = -sqrt(r^2 - xt^2)
and the angle you mention would be
θ' = arccos(|xt| / r)

For your example picture you would set θ = -30°, for example.

[Edited by - mattd on October 27, 2009 12:04:16 AM]

##### Share on other sites
Sorry, I'm very new to this kind of math. ( I'm self taught ) This is because I'm still in high school and we haven't covered anything like this, I've just gone through algebra and some geometry.

So what exactly is θ' = arccos(|xt| / r) I understand arccos, radius but whats Xt? Just simply the x coord? And xt = r.tan(θ) whats the r infront of tan and what do you plug in for θ? I know its an angle but would it be the 30 degrees that I had in my example or the one we get from θ' = arccos(|xt| / r) Sorry, I'm a noob :P

Well anyway, thanks for the reply, I've been confused on this all week and I'm hoping I can put an end to this.

##### Share on other sites
θ = input angle
Xt = target point's X coordinate
Yt = target point's Y coordinate
θ' = second angle that you mentioned you were looking for

Quote:
 Original post by zaneski13So what exactly is θ' = arccos(|xt| / r) I understand arccos, radius but whats Xt? Just simply the x coord?

Yup, and the formula for figuring it is the first one, Xt = r.tan(&theta)

Quote:
 And xt = r.tan(θ) whats the r infront of tan

The circle's radius, same as you mentioned above. The dot just means multiply.

Quote:
 and what do you plug in for θ? I know its an angle but would it be the 30 degrees that I had in my example or the one we get from θ' = arccos(|xt| / r)

I'm guessing that the grey solid line extending from the camera point (at the top of the circle) is the camera's view direction. With that in mind, θ is the anticlockwise angle of your camera view direction from the y-axis.
Check out your example picture. Here you'd set θ = -30°, because the camera is looking 30° to its right. If it was looking towards its left, it would instead be +30°.

θ' (this is different from your input angle θ!) is the angle you specified on your diagram as "We need to find this angle", in that lime colour.

I put the equations in order so calculating the first gives a result you need for the second.
So you have start with your known θ, and find out Xt. Then you use that to find Yt, and θ' (assuming you still need it.. if all you needed was the location of the target point T, you have it already at (Xt, Yt).

[Edited by - mattd on October 27, 2009 10:34:53 PM]

1. 1
2. 2
3. 3
Rutin
14
4. 4
5. 5

• 9
• 9
• 11
• 11
• 23
• ### Forum Statistics

• Total Topics
633674
• Total Posts
3013277
×