त्रिभुज ABC का परिकेन्द्र O हे, यदि `angleBAC=50^(@)` हे, तो `angleOBC=?`

Options

A. `100^(@)`
B. `130^(@)`
C. `40^(@)`
D. `50^(@)`

Answer

C

Explanation

`therefore angleBAC = 50^(@)` दिया गया हे
जैसा हम जानते हे
`rArr angleBOC = 50^(@)xx2` (with chord BC)
`rArr OB=OC=r`
Then तब `angleOBC =angleOCB = x^(@)`
`rArr angleOBC + angleOCB + angleBOC= 180^(@)`
`rArr anglex^(@) + x^(@) + 100 = 180^(@)`
`rArr x=40^(@)`
Therefore (अन्तः) `angleOBC = 40^(@)`