How find the angle subtended by a chord at the circumference of a circle

C

chijioke

Full Member
Joined
Jul 27, 2022
Messages
352
A chord is 3 cm from the center of a circle of radius 14 cm. Calculate the angle subtended by the chord at the circumference of the circle?
I want solve this problem. But at first I want to know the relationship between the angle subtended by a chord at the circumference of a circle and the angle subtended by the same chord at the center of the same circle. I don't know does any such relationship exist?
 
chijioke said:
From the diagram. I can see that the angle subtended by chord at the center of the circle is [math]2\theta[/math] but what if the same chord also subtended another angle p, at circumference as in:View attachment 34929
Click to expand...
Post #3 answers your question directly! Did you not see it?
If you want further confirmation have a look at this site (read down at least as far the "Angle at the Centre Theorem" but you should study & learn everything on that page).

PS:- (Later Edit)
Important Note: You have marked the "3cm" in the wrong place on your diagram (above); compare your diagram with that shown at Post #2 (which is the correct location for the "3cm" distance that is needed to calculate the "θ" shown in that diagram)!
 
Last edited:
Hi @chijioke,

You were in the forum when I posted my Reply (above) to your Post (#4) but I later added an Important Note to my post that you may not have seen before you left the forum.

You should go back into the forum and read what I added before attempting to complete your answer to the problem you were given.

Please then show us your answer so that we can con firm that it is correct (or offer further advice).

Regards.
 
The Highlander said:
Post #3 answers your question directly! Did you not see it?
If you want further confirmation have a look at this site (read down at least as far the "Angle at the Centre Theorem" but you should study & learn everything on that page).

PS:- (Later Edit)
Important Note: You have marked the "3cm" in the wrong place on your diagram (above); compare your diagram with that shown at Post #2 (which is the correct location for the "3cm" distance that is needed to calculate the "θ" shown in that diagram)!
Click to expand...
IMG_20230201_103342.jpg
Sorry, that is an oversight. From my diagram above, you can see that chord AB subtends central angle [math]2\theta[/math] as well as angle P at point C in the circumference? My question is what is the relationship between central angle [math]2\theta[/math] and angle p at the circumference ?
 
Harry_the_cat said:
View attachment 34926
Click to expand...
The diagram is showing that the arc AB subtends central angle 2x and angle x at the circumference at point C.
From the diagram, I can see that central angle is twice the angle at the circumference.
Arc and chord are not the same. What happens when a chord subtends an angle at the center of a circle and also subtends another at the circumference as in my diagram in post 7?
 
Last edited:
chijioke said:
The diagram is showing that the arc AB subtends central angle 2x and angle x at the circumference at point C.
From the diagram, I can see that central angle is twice the angle at the circumference.
Arc and chord are not the same. What happens when a chord subtends an angle at the center of a circle and also subtends another at the circumference as in my diagram in post 7?
Click to expand...
You are correct in saying that a chord and an arc are not the same thing.
BUT, the angle subtended by a chord at the centre of a circle is the same as the angle subtended by an arc at the centre of a circle.
1675249911561.png
I've drawn in the chord AB on the diagram I posted in Post #3.
The same relationship between the angles holds.

I am not quite sure why you are not understanding that.

1675250091824.png
Using your diagram, can you see that the magnitude of <AOB is twice the magnitude of <ACB ?
 
Harry_the_cat said:
You are correct in saying that a chord and an arc are not the same thing.
BUT, the angle subtended by a chord at the centre of a circle is the same as the angle subtended by an arc at the centre of a circle.
View attachment 34932
I've drawn in the chord AB on the diagram I posted in Post #3.
The same relationship between the angles holds.
Click to expand...
I was thinking the same initially but I don't want to jump into conclusion. Now, I get it. Expect my answer soon.
Harry_the_cat said:
I am not quite sure why you are not understanding that.

View attachment 34933
Using your diagram, can you see that the magnitude of <AOB is twice the magnitude of <ACB ?
Click to expand...
Yes I can see it.
 
The Highlander said:
Hi @chijioke,

You were in the forum when I posted my Reply (above) to your Post (#4) but I later added an Important Note to my post that you may not have seen before you left the forum.

You should go back into the forum and read what I added before attempting to complete your answer to the problem you were given.

Please then show us your answer so that we can con firm that it is correct (or offer further advice).

Regards.
Click to expand...
IMG_20230201_132848.jpg
Angle at circumference is 77.54 degrees.
 
Well done. A couple of things to make it even better:
1. I think you meant to write \(\displaystyle 3^2\) where you have \(\displaystyle 9^2\). You seem to have used \(\displaystyle 3^2\) in your calculation.
2. \(\displaystyle \sqrt{14^2-3^2}\) is positive only.
3. Because you have rounded 13.67 to 2 decimal places this introduces an error into your final answer.
4. It would have been easier to use \(\displaystyle cos\theta=\frac{3}{14}\) to find \(\displaystyle \theta\). Saves using Pythagaras' theorem and will give a more precise answer (approx 77.63 degrees).
 
Harry_the_cat said:
Well done. A couple of things to make it even better:
1. I think you meant to write \(\displaystyle 3^2\) where you have \(\displaystyle 9^2\). You seem to have used \(\displaystyle 3^2\) in your calculation.
2. \(\displaystyle \sqrt{14^2-3^2}\) is positive only.
3. Because you have rounded 13.67 to 2 decimal places this introduces an error into your final answer.
4. It would have been easier to use \(\displaystyle cos\theta=\frac{3}{14}\) to find \(\displaystyle \theta\). Saves using Pythagaras' theorem and will give a more precise answer (approx 77.63 degrees).
Click to expand...
Thanks for the observations. But it amazes me how you got to know that 77.54 degrees is not exact. At first sighting what did you think or see that made you see that my solution is not exact?
 
Last edited:
chijioke said:
Thanks for the observations. But it amazes me how you got to know that 77.54 degrees is not exact. At first sighting what did you think or see that made you see that my solution is not exact?
Click to expand...
The cat would have immediately noted your rounding (to "13.67") and almost certainly, therefore (just to check it), have plugged in the simple calculation: 3 ÷ 14 and found the Arccosine of the answer (without any rounding) to get 77.62637.... ≈ 77.63°.

That's what I (& anyone following sound Maths practice) would have done.

It's worth noting too, now that you understand that an arc & its chord subtend the same angle at the centre and have now learned all about angles inscribed in circles (by studying the page I sent you to ?), that, in your diagram, you could just have named the angle "P" as "θ" since it is the same as the angle "θ" you've shown at the centre in your sketch. And, since you (now) have the hypotenuse and the side adjacent to "θ" in you sketch, you can go straight to use of the Cosine Ratio to find the measure of "θ". (No need to mess about with Pythagoras or Arcsine or finding the measure of the angle at the centre only to half it again! ?)

I trust you have learned quite a bit from this little exercise and the help that has been offered in the forum and wish you 'All the Best' in your future Maths studies but perhaps the most important "moral of the story" is:-

Always leave rounding to your final answer! (Wherever possible, of course. ?)
(Make good use of your calculator's memory function. ?)​
 
The Highlander said:
The cat would have immediately noted your rounding (to "13.67") and almost certainly, therefore (just to check it), have plugged in the simple calculation: 3 ÷ 14 and found the Arccosine of the answer (without any rounding) to get 77.62637.... ≈ 77.63°.

That's what I (& anyone following sound Maths practice) would have done.

It's worth noting too, now that you understand that an arc & its cord subtend the same angle at the centre and have now learned all about angles inscribed in circles (by studying the page I sent you to ?), that, in your diagram, you could just have named the angle "P" as "θ" since it is the same as the angle "θ" you've shown at the centre in your sketch. And, since you (now) have the hypotenuse and the side adjacent to "θ" in you sketch, you can go straight to use of the Cosine Ratio to find the measure of "θ". (No need to mess about with Pythagoras or Arcsine or finding the measure of the angle at the centre only to half it again! ?)

I trust you have learned quite a bit from this little exercise and the help that has been offered in the forum and wish you 'All the Best' in your future Maths studies but perhaps the most important "moral of the story" is:-
Click to expand...
Noted!
The Highlander said:
Always leave rounding to your final answer! (Wherever possible, of course. ?)
(Make good use of your calculator's memory function. ?)​
Click to expand...
Noted!
 
You must log in or register to reply here.
Top