Angle between a chord and a tangent

[Gorgiewave]

Please see the attached sketch. CB is a chord. D is the midpoint of the line CB. The other line is a tangent to the circle that is parallel to the chord CB. I'm looking for a way to calculate the angle CEB. Can anyone explain how that could be done? I think if the chord was a diameter line, CEB would be 90º (π/2 rad) and if the chord were overlaid on the tangent, CEB would be 180º (π rad).

 

[Dr.Peterson]

Angle CEB is half of angle CAB (on the side away from E). This is because of a theorem that an inscribed angle is half of the arc it intercepts. In fact, this is true for any point E on the arc CB, not just the midpoint.

In your example of a diameter, 90 is half of 180.

 

[Gorgiewave]

PS: I might add that this is not a homework question, just something I've been trying unsuccessfully to work out.

If anyone knows the function of how the angle changes as the chord moves (lengthening until it becomes a diameter line, then shortening again) while remaining parallel to the tangent, I'd be grateful.

 

[Gorgiewave]

Angle CEB is half of angle CAB (on the side away from E). This is because of a theorem that an inscribed angle is half of the arc it intercepts. In fact, this is true for any point E on the arc CB, not just the midpoint.

In your example of a diameter, 90 is half of 180.

Many thanks for that. I'm going to look at that to try to understand it properly.

 

[Dr.Peterson]

PS: I might add that this is not a homework question, just something I've been trying unsuccessfully to work out.

If anyone knows the function of how the angle changes as the chord moves (lengthening until it becomes a diameter line, then shortening again) while remaining parallel to the tangent, I'd be grateful.

What are the inputs to the function you have in mind? Fixed radius, and variable distance from the center, or from the fixed tangent, I suppose?

See here for many formulas: https://mathworld.wolfram.com/CircularSegment.html

As for the angles: https://mathbitsnotebook.com/Geometry/Circles/CRAngles.html

 

[Gorgiewave]

What are the inputs to the function you have in mind? Fixed radius, and variable distance from the center, or from the fixed tangent, I suppose?

See here for many formulas: https://mathworld.wolfram.com/CircularSegment.html

As for the angles: https://mathbitsnotebook.com/Geometry/Circles/CRAngles.html

Yes, I'm picturing the tangent remaining in place and the chord moving up and down, always remaining parallel to the tangent. At CAB = 0º, CB would have zero length and at CAB = 180º, CB would have length D. I'm looking for the relationship between angle CAB and the length of the chord.

Chord length appears to go in a 0 - D - 0 cycle while CAB goes in a 0 - 180 - 0 or 0 - 360 cycle.

What does "on the side away from it" mean in "Angle CEB is half of angle CAB (on the side away from E)."? In the sketch I posted, angle CEB looks clearly larger than angle CAB.

 

[Gorgiewave]

I've now discovered that a kite is a right kite if it can be inscribed in a circle. I would now solve this as: total internal angle of kite = 360º. GBF and GCF are 90º = 180º. BGC = 1/2 BAC (per inscribed angle theorem), hence BFG = 180 - 1/2 BAC. Please tell me if this reasoning would give a correct value for BFG (if any actual values were known).

 

[pka]

Using your diagram above in reply #7.
\(m(\widehat{ BFC})=m(\angle BAC)\) also \(m(\angle BGC)=\frac{1}{2}m(\widehat{ BFC})\)
\(m(\angle BGC)=\frac{1}{2}m(\widehat{ BFC})\)
m is for measure.

 

[Gorgiewave]

Using your diagram above in reply #7.
\(m(\widehat{ BFC})=m(\angle BAC)\) also \(m(\angle BGC)=\frac{1}{2}m(\widehat{ BFC})\)
\(m(\angle BFC)=\frac{1}{2}m(\widehat{ BFC})=\frac{1}{2}m(\angle BAC)\)

Thank you very much. I don't know what the m refers to.

∠BFC looks larger than ∠BAC, yet according to your last line it is half of ∠BAC. Please explain how this is so or what specifically I have missed.

 

[Dr.Peterson]

Yes, I'm picturing the tangent remaining in place and the chord moving up and down, always remaining parallel to the tangent. At CAB = 0º, CB would have zero length and at CAB = 180º, CB would have length D. I'm looking for the relationship between angle CAB and the length of the chord.

Chord length appears to go in a 0 - D - 0 cycle while CAB goes in a 0 - 180 - 0 or 0 - 360 cycle.

What does "on the side away from it" mean in "Angle CEB is half of angle CAB (on the side away from E)."? In the sketch I posted, angle CEB looks clearly larger than angle CAB.

The rays AB and AC form two angles, one above and one below. Both can be called angle CAB; one (above) is what you are seeing, and the other (below) is what I called "on the side away from E" -- the proper term is "reflex angle", meaning it is greater than 180 degrees.

How much trigonometry do you know? The facts you've stated are correct, but you can get your answer without them.

 

[Gorgiewave]

The rays AB and AC form two angles, one above and one below. Both can be called angle CAB; one (above) is what you are seeing, and the other (below) is what I called "on the side away from E" -- the proper term is "reflex angle", meaning it is greater than 180 degrees.

How much trigonometry do you know? The facts you've stated are correct, but you can get your answer without them.

Thank you. I know basic or school trigonometry. I don't know what specifically would be helpful here.

Let's assume we measured ∠BAC in the second sketch and it is 80º (a reasonable estimate by sight). What would ∠BFC be? (360-80)/2 = 140º?

 

[Dr.Peterson]

Here is my picture, with the parts you need to find angles CAB and CEB. (You've mentioned wanting both.)



Angle BAC is twice angle EAC; you know r and c, so you can find the angle using an inverse sine.

To find angle BEC from that, you can use the facts you've found. Or, use the Pythagorean theorem to find d, use that to find h, and then use an inverse tangent to find angle AEC.

 

[Gorgiewave]

Here is my picture, with the parts you need to find angles CAB and CEB. (You've mentioned wanting both.)

View attachment 19822

Angle BAC is twice angle EAC; you know r and c, so you can find the angle using an inverse sine.

To find angle BEC from that, you can use the facts you've found. Or, use the Pythagorean theorem to find d, use that to find h, and then use an inverse tangent to find angle AEC.

Thank you. I know enough trigonometry to follow that. I'm grateful for your patience.

 

[Dr.Peterson]

Let's assume we measured ∠BAC in the second sketch and it is 80º (a reasonable estimate by sight). What would ∠BFC be? (360-80)/2 = 140º?

The reflex angle ∠BAC would be 360º - 80º = 280º; so ∠BFC would be half of that, 140º. You're correct.

Alternatively, ∠BGC = ∠BAC/2 = 40º, so∠BFC is the supplement of that, 140º.

 

[Gorgiewave]

The reflex angle ∠BAC would be 360º - 80º = 280º; so ∠BFC would be half of that, 140º. You're correct.

Alternatively, ∠BGC = ∠BAC/2 = 40º, so∠BFC is the supplement of that, 140º.

Thank you. I'm going to try and digest all this.