Question

In the given figure, O is the centre of a circle and arc AB = arc BC = arc CD. If ∠COB = 60˚, find: i) ∠ AOC ii) ∠DOC iii) ∠OAC (Give Reasons) *please with full explanation

Answer 1

Answer:

angle AOC= 120°

angle DOC =60°

Angle OAC = 30°

first of all there is a formula for arc of circle

it is length of arc = angle multiple by radius

so let the arc AB=Arc BC = Arc CD = X

and radius be r

so by formula

BC = 60* r - 1

BC/r = 60. -3

AB = angle AOB* r. -2

Adding 1 and 2

we get

AB+BC = (60 + AOB) * r

But AB and BC are equal

so 2BC = (60 + AOB)*r

so 2 BC/r = ( 60+ AOB). -4

but by 3

we know that BC/r = 60

so putting it in 4

we get

2 * 60 = AOB +60

120-60 = AOB

AOB =. 60°

AOC = AOB +BOC

AOB= 60°& BOC =60°

therefore angle AOC= 120°

For angle DOC it is the same process

For OAC

first connect A and C

OA and OC are equal [ Because the are radius of circle]

so it is and isosceles triangle

so angle OAC and angle OCA are equal

in triangle sum of all angle are 180°

so angle AOC + OAC +OCA =180°

but we know that AOC=120 and OAC and OCA are equal

so 2 OAC +120° = 180°

2 OAC = 60

OAC = 30 °

Hope it help u

Answer 2

Step-by-step explanation:

i) arc BC=arc AB....

so, AOB = BOC

AOB=60°

SO, AOC=AOB+BOC

AOC=60+60

120°

ii) arc DC = arc BC

so, DOC=COB

DOC=60°

iii) First join AC

OA=OC (radius)

so,in this triangle OAC,two sides are equal,so it is an isosceles triangle. That means two angles are also same......

AOC=120°

we,know that sum of three interior angles of a triangle is 180°

so, AOC+OAC+OCA=180°

120+2OAC=180° (as OAC and OCA are

same so i have written

2OAC)

2OAC=180-120

2OAC=60

OAC=60/2

so, OAC=30°

U can see the picture for question iii .

hope it helps.........

thank u..........