Question

solve this please I need it urgently tomorrow is my exam​

Answer 1

Step-by-step explanation:

• <AOB = 104°
• Arc AB = 2*Arc BC

NOW,

WE KNOW THAT EQUAL ARC SUBSTEND EQUAL ANGLES AT THE CENTER OF THE CIRCLE

SO,

ARC AB = 2*ARC BC

=> <AOB = 2*<BOC

=> 104°/2 = <BOC

=> <BOC = 52°

_______________________

NOW,

IN ∆ AOC,

OA = OC [RADIUS OF THE SAME CIRCLE]

SO,

<OAC = <OCA [ANGLES OPPOSITE TO THE EQUAL SIDES ARE EQUAL IN A TRIANGLE]

<AOC = <AOB + <BOC = 104° + 52° = 156°

THEREFORE,

<AOC + <OAC + <OCA = 180° [ SUM OF ALL INTERIOR ANGLES IN A TRIANGLE]

=> 156° + <OAC + < OAC = 180°. [<OCA = <OAC]

=> 2<OAC = 28°

=> <OAC = 14°

___________________________

<BAC = (1/2)<BOC. [ ANGLES BY THE ARC AT THE CENTER IS DOUBLE THE ANGLE SUBTENDED AT THE ANY PART OF THE CIRCLE]

<BAC = (1/2)* 52 = 26°

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//HOPE THIS WILL HELPED YOU//

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