`DeltaABC` में O परिकेन्द्र हे। यदि `angleBAC=85^(@)` और `angleBCA=75^(@)` हे, तो `angleOAC` किसके बराबर होगा?

Options

A. `70^(@)`
B. `60^(@)`
C. `50^(@)`
D. `40^(@)`

Answer

A

Explanation

`therefore` O is circumcentre
So, OA =OB=OC=R (Radius)
`therefore` `[angleABC]=180^(@)-(angleBAC + angleBCA)`
`angleABC = 180^(@)-(85^(@)+75^(@)`
`angleABC = 180^(@)-160^(@)`
`rArr angleABC = 20^(@)`
Angles made by same chord at the center is doubled than that of any other part of the circumference at same sector)
Then, `angleAOC = 20^(@)xx 2=40^(@)`
`therefore` OA =OC=R
So, `DeltaAOC` = Isosceles triangle
Then, `angleOCA + angleOAC + angleAOC = 180^(@)`
`2x+40^(@)=180^(@)`
x=`70^(@)`
therefore , `angleOAC = 70^(@)`