Question

in fig. 16.201,ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that angle AED=95°and angle OBA=30°. Find angle OAC

Answer 1

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° ANS

Answer 2

Replace the <AED =95° with <ADE=95°.

Because this is an error printing in question.

Step by Step Solution :

<ADE and <ADC forms a linear pair .

So,

<ADE+<ADC= 180°

95°+<ADC=180(given <ADE=95°)

<ADC=180°-95°

<ADC=85°

Now see that the same arc AC made an angle at circumference of circle i.e.,<ADC and another angle <AOC at the centre of the circle .

So

<AOC=2<ADC

<AOC=2×85°

<AOC=170°

Now in ∆AOC

OA =OC (radii of the circle)

So <OAC = <OCA

Now

<OAC+<OCA+<AOC=180°

<OAC+<OAC+170°=180°

2<OAC=180°-170°

2<OAC=10°

<OAC=10°/2

So <OAC =5°