Question

In the given figure, 'O' is the centre of the circle in which ZOBA = 20° and ZOCA = 30° then BAC​

Answer 1

Answer:

in triangle AOB

AO=BO(radii of the same circle)

therefore :<OBA=<OAB

20=<OAB. (1)

in triangle AOC

AO=CO(radii of the same circle)

therefore :<OCA=<OAC

30=<OAC. (2)

by (1) and(2)

<OAB+<OAC=<BAC

20+30=<BAC

50=<BAC

hope it was help ful

Answer 2

Answer:

Hay Guys!!!!

Step-by-step explanation:

We have OB=OA(radii of the same circle)

Therefore OAB is an issosceles triangle

Angle OBA =AngleOAB

In Triangle OAB

Angle OAB =20 degree

Similarly in triangle OAC

Angle OAC =30 degree

Total angle BAC =50 degree

We know, BAC =halfBOC

Therefore angle boc =100 degree

Hence founded

Hope it may help to you