# find arc BN=?

#### Albert

##### Well-known member
Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on

segment MN , if :

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

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#### HallsofIvy

##### Well-known member
MHB Math Helper
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?

#### Albert

##### Well-known member
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.

#### Opalg

##### MHB Oldtimer
Staff member
Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on

segment MN , if :

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

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 $\overset{\frown} {BN}= \angle BCN = 60^\circ.$

#### Albert

##### Well-known member
what will be the value of arc BN , if point P locates between points C and N

#### Opalg

##### MHB Oldtimer
Staff member
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 $\overset{\frown} {BN} = 20^\circ.$

#### Albert

##### Well-known member
very good solution
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