area of circle

Question

In a circle , AB=AC are two chords. If ∠BOC= 120°, find ∠BAC, ∠ABO and ∠ACO

Answer 1

From the problem statement, it is assumed that the center of the 

circle is point "O".  It helps if you draw a diagram and label the

given info and info as we as find.  As we discuss the solution,

trace each bit of info on your diagram to identify the item discussed.  

All numbers shown are in degrees.  From given info, we can say...

 

∠BOC=120 is a central angle which is equal to arc BC

∠BAC is an inscribed angle

∠BAC=(1/2)arc BC....inscribed angle is 1/2 of arc length (or

                                central angle measure)

        =(1/2)(120)

        =60

 

Now draw line segments OA, OC, OB.  These segments are all 

radii of circle "O," all having the same length.  We have now 

two isosceles Δ's ABO, ACO.

 

        360-∠BOC=∠BOA+∠AOC
           360-120=∠BOA+∠AOC
∴ ∠BOA+∠AOC=240

 

Segment OA bisects ∠BOA+∠AOC, so...

∠BOA=(1/2)(240)

∠BOC=120...and...

∠AOC=120

 

With isosceles Δ's, we know that base angles are equal, so...

∠ABO=∠BAO=∠ACO=∠AOC=(180-120)/2

                                         =60/2

                                         =30

 

 

 

Answer 2

∠BOC is a central angle, therefore arc BC = 120°.

∠BAC is an inscribed angles, therefore ∠BAC = 60°

∠OAC = 30°, therefore ∠ACO = 30°

The same is true for ∠ABO, it equals 30°