Question

in the figure ABCD is a quadrilateral inscribed in a circle with Centre O CD is produced to e if angle a b is equal to 70 degree and Angle A b is equal to 45 degree calculate angle AOC and angle BAC.​

Answer 1

where is the figure!

Answer 2

Here AED will not be 95° , ADE is 95°
solution ÷ Using linear pair therom,
ADE + ADC = 180°
or, ADC = 180°- ADE = 180°-95°
or, ADC = 85°
Now, because ABCD is a cyclic quadrilateral
so,
ABC + ADC= 180°
or, (ABO +OBC)+ADC= 180°
or, ( 30° +OBC) +95°= 180°
or, OBC= 180°- 125°= 65°

As, BO and CO are radii of same circle
so, BO=CO
so, OBC=OCB [ angle opposite to equal
opposite sides are equal]
In Triangle OBC , by sum property

OBC+ OCB + BOC=180°
by putting all the values
BOC= 50°
or, 2BAC = 50° [we know angle
substanded by an arc
at the centre is twice
the angle substanded
by it at remaining circle]
or, BAC=25°

Again,
In Triangle AOB
AO= BO [ Radii of same circle]
so, OAB=OBA (=30°)

Finally,
OAC= OAB - BAC= 30°-25°=5°
Hence,
OAC=5°