Finding $\overset{\frown}{BN}$ with Given Parameters

  • Context: MHB 
  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Parameters
Click For Summary

Discussion Overview

The discussion revolves around finding the measure of arc $\overset{\frown}{BN}$ in a geometric configuration involving a circle, points on the circle, and angles formed by those points. The problem is framed in the context of a circle with a diameter and explores various placements of points within that configuration.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • Post 1 presents the problem of finding $\overset{\frown}{BN}$ given specific angles and arc measures.
  • Post 2 and Post 3 clarify the terminology, confirming that "C" refers to the center of the circle.
  • Post 4 proposes a solution where if point B is opposite A, then $\overset{\frown}{BN} = 140^\circ$, but suggests that the more interesting case is when A and B are on the same side of MN, leading to a calculation of $\overset{\frown}{BN} = 60^\circ$.
  • Post 5 and Post 6 inquire about the value of arc BN if point P is located between points C and N, suggesting that this changes the angle ABC and leads to a different calculation resulting in $\overset{\frown}{BN} = 20^\circ$.
  • Post 7 acknowledges the variability of the solution based on the positions of points B and P, emphasizing that the problem can yield multiple answers depending on the diagram's configuration.

Areas of Agreement / Disagreement

Participants express different approaches to the problem, with some proposing specific solutions based on varying placements of points. There is no consensus on a single solution, as the discussion highlights multiple scenarios leading to different arc measures.

Contextual Notes

The discussion reveals dependencies on the placement of points and the assumptions made about angles and arcs, which may affect the outcomes. The problem's open-ended nature allows for various interpretations and solutions.

Who May Find This Useful

Readers interested in geometric problems, angle relationships in circles, and those exploring open-ended mathematical challenges may find this discussion relevant.

Albert1
Messages
1,221
Reaction score
0
Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on

segment MN , if :

$\angle CAP=\angle CBP =10^o ,\,\, \overset{\frown} {MA}=40^o,\,\, find :\,\, \overset{\frown} {BN}=?$
 
Last edited:
Mathematics news on Phys.org
You refer to "circle C" but then treat "C" as if it were a point. Are we to assume that "C" is the center point of the circle?
 
HallsofIvy said:
You refer to "circle C" but then treat "C" as if it were a point. Are we to assume that "C" is the center point of the circle?

yes ,you got it !
"C" is the center point of the circle.
 
Albert said:
Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on

segment MN , if :

$\angle CAP=\angle CBP =10^o ,\,\, \overset{\frown} {MA}=40^o,\,\, find :\,\, \overset{\frown} {BN}=?$
[sp]

One solution is for $B$ to be opposite $A$ on the other side of $MN$, at the point labelled $B'$ in the picture. Then $\overset{\frown} {BN} = 140^\circ$. But that is too obvious to be interesting, and I assume that what was wanted is the case where $A$ and $B$ are on the same side of $MN$.

The points $A, B, C, P$ are concyclic, because $\angle CAP=\angle CBP =10^\circ$. Therefore $\angle ABP=\angle ACP =40^\circ$, and so $\angle ABC= 10^\circ + 40^\circ = 50^\circ.$ The triangle $ABC$ is isosceles, so $\angle BAC = 50^\circ$, and $\angle ACB =80^\circ$. Finally, $\angle BCP =40^\circ + 80^\circ = 120^\circ$, from which $\overset{\frown} {BN}= \angle BCN = 60^\circ.$[/sp]
 

Attachments

  • angles.png
    angles.png
    6.9 KB · Views: 96
what will be the value of arc BN , if point P locates between points C and N
 
Albert said:
what will be the value of arc BN , if point P locates between points C and N
Good question! I hadn't thought of that possibility. The method will be similar to the previous one, but this time the angle ABC ($\angle A'B'C$ in the diagram below) will be $40^\circ - 10^\circ = 30^\circ$ instead of $40^\circ + 10^\circ = 50^\circ$. Then $\angle A'CB' = 120^\circ$ and $\overset{\frown} {BN} = 20^\circ.$
 

Attachments

  • angles2.png
    angles2.png
    9.6 KB · Views: 97
very good solution :cool:
this is an open -style problem ,if the position of point B or point P changes ,then the answer will also differ (it depends upon how the diagram is sketched)
sometime we may give students a mathematic problem with more then one possible answer
 

Similar threads

MHB Solve Exercise 1.6 in Greenberg's Euclidean Geometry: Betweenness and Lying On
  • · Replies 1 ·
Replies
1
Views
3K
MHB Calculate the radius of the circle
  • · Replies 5 ·
Replies
5
Views
3K
MHB Find the area of the shaded region
  • · Replies 2 ·
Replies
2
Views
1K
MHB State the angles in terms of x , Circle theorems
  • · Replies 6 ·
Replies
6
Views
2K
MHB Find the square of the distance
  • · Replies 1 ·
Replies
1
Views
1K
MHB Find the Bearing from A to C & Angle B: Solve Here
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad A check: shortest distance from point to line
  • · Replies 2 ·
Replies
2
Views
2K
MHB What is the value of Angle BMC in Triangle ABC?
  • · Replies 1 ·
Replies
1
Views
2K
MHB Find the angle, cyclic quadrilaterals
  • · Replies 8 ·
Replies
8
Views
3K
MHB Solving Acute Triangle Angles Given $A,B,C$ & Equations
  • · Replies 1 ·
Replies
1
Views
2K