Question

I want the answer of these three questions
Step by stem plzzzzzzz
As fast as you can

Answer 1

Question 16-

in ∆ACD,
1 + 2 = 180° - 110°
=> [1 + 2= 70°]

ADE = ABC. [Ext. angle of cyclic quadrilateral]
=> ABC = 70° = OBA + OBC
=>OBC =70°- OBA =70° - 45°

[OBC = 25°]

In ∆OBC,
OB = OC [radii]
=> 3 + 4 = 25° [angle opposite to equal sides]

In Cyclic Quadrilateral ABCD,

A+C = 180° [Opposite angles of cyclic quadrilateral]

(45°+X+2) + (25°+X+1) = 180°
70° + 2X + (1+2) = 180°
2X = 180° - 70° - 70°. [1+2 = 70°]
2X = 40°
X=20°

=> [OCA= 20°]

BAC = 45° + X
= 45° + 20°

[BAC= 65°]