How to Calculate Tangent Lines Given Angle and Radius of Circle/Arc

[JDfromtheIE]

This is my first post here. Thank you for Having me.

I need to figure out a formula to calculate the length of 2 tangent lines given the radius of the circle/arc and the angle of the 2 lines.

A little background on what I'm trying to achieve...

I'm an architectural woodworking engineer (fancy for cabinet engineer). I'm trying to add a radius to what has always been a straight angled notch.

This radius will always be 40mm but the angle between these 2 tangent lines can and will change. This angle will always be obtuse do to the design of this cabinet.

I've attached 1 image showing how I calculate the angles and the other image shows my question.

Any help would be greatly appreciated

 

[LCKurtz]

Go to this link. I have set it up so you can move point B around for different angles.

Geometry - GeoGebra

Interactive, free online geometry tool from GeoGebra: create triangles, circles, angles, transformations and much more!

www.geogebra.org

 

[JDfromtheIE]

Thank you. That's extremely helpful. How do I write that into a formula though?

 

[LCKurtz]

You don't need to. Just save the link and use it whenever you need it. Why re-invent the wheel?

 

[Mr. Bland]

Got you covered, @JDfromtheIE. Consider a more general form of this problem as depicted by this crappy drawing:

​

Imagine this is to scale and the points on the circle where the dotted lines meet form right angles with each other. Fundamentally what you need to do is give some angle for [MATH]\theta[/MATH] and calculate [MATH]L[/MATH] from that.

Angle [MATH]q[/MATH] works out to be the opposite angle from [MATH]\theta[/MATH], so its measure is [MATH]q = 180^\circ - \theta[/MATH]. Connecting the center of the circle with the "tip" of the projection, we form two mirrored right triangles with angles [MATH]90^\circ[/MATH], [MATH]\frac{\theta}{2}[/MATH] and [MATH]\frac{q}{2}[/MATH], and one of the side lengths is [MATH]r[/MATH]. We don't know the other side lengths, but we can get there from here.

The law of sines gives a convenient relationship: dividing a triangle's side length by the sine of the opposite angle gives us the same ratio for all three side-angle pairs in any triangle. To solve for [MATH]L[/MATH], we can make the following observation:

[MATH]\frac{L}{sin\left(\frac{q}{2}\right)} = \frac{r}{sin \left(\frac{\theta}{2}\right)}[/MATH]
[MATH]L = \frac{sin\left(\frac{q}{2}\right) * r}{sin \left(\frac{\theta}{2}\right)}[/MATH]​

Substitute the expression for [MATH]q[/MATH]:

[MATH]L = \frac{sin\left(\frac{180^\circ - \theta}{2}\right) * r}{sin \left(\frac{\theta}{2}\right)} = \frac{sin\left(90^\circ - \frac{\theta}{2}\right) * r}{sin \left(\frac{\theta}{2}\right)}[/MATH]​

And there we have it. The radius of the circle in the original post is [MATH]r = 40[/MATH]mm. Given the angle for [MATH]\theta[/MATH], the length of the tangent lines can be found with the above formula.

Why re-invent the wheel?

Because that's how we do. Why depend on a calculator that requires internet access when one can understand how to solve the problem?

 

[Mr. Bland]

Derp, had tunnel vision and didn't realize that [MATH]sin(90^\circ - x) = cos(x)[/MATH]. The formula further simplifies to this:

[MATH]L = \frac{cos\left(\frac{\theta}{2}\right) * r}{sin\left(\frac{\theta}{2}\right)}[/MATH]​

__________

EDIT:
Seems this even further simplifies to [MATH]L = cot\left(\frac{\theta}{2}\right) * r[/MATH]. I guess I need to go back to school.

 

[JDfromtheIE]

Thank you so much @Mr. Bland

This is exactly what I needed. I have to write a parametric program / formula for this cabinet product so that I can adjust the variables like ADA height, depth and Toe Space Depth so that the radius moves where it needs to be.

This helps tremendously

 

[JDfromtheIE]

So I'm having a hard time getting this to evaluate in excel...

The program I use to create these "Products" is excel based.

θ=99.5 and r=40 for this specific equation

In excel it would be... "=cot( 99.5/2)*40" correct?

 

[JDfromtheIE]

Nevermind... I needed to convert the angles from radians to decimal

 

[Deleted member 4993]

Nevermind... I needed to convert the angles from radians to decimal

Please look at the user' guide for MS-excel very carefully. As far as I know, the input for angle measures in trigonometric functions while using excel

Must be in radians

You can check that out by evaluating sin(Θ) or cos(Θ) or tan(Θ) for 30^o and π/6 radians.

 

[JDfromtheIE]

I just wanted to thank you guys again...

The product is working perfectly and I couldn't have done it without you guys' help.

It's completely parametric and works exactly as intended

 

[Mr. Bland]

Splendid! It's always exciting to see how mathematics are applied in the real world.