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Angle AED= 47 ( Alternate angles are equal)
Angle BCD= 124 ( Cyclic Quadrilateral, opposite angle add up to 180)
Angle ADB= 47 ( Alternate angles are equal)

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aed=47 alt angles
bcd=124 cyclic quad
adb=47 alt seg

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Q2) AED=47 ( alt angle)
BCD=180-56=124 (cyclic quadrilateral )

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Munahil=BCD=124(quadrilateral cyclic theorim)

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AED= 47
As all alternate angles are equal
BCD= 180-56= 124
(Cyclic quadrilateral theorem)
ADB= 47
(Alternate segment theorem)

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Angle AED makes an alternate angle so its 47
Angles BCD is 124 as it makes the cyclic quadrilateral with 56 on the opposite side so 180-56= 124

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AED=47 (alternate angles)

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Munahil-AED=47( the two lines are parallel and they make a Z)

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Q2) angle AED = 47 because alternate angle
Angle BCD = 180 - 56 = 124 because cyclic quadrilateral

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BCD = 180-56 = 124 (cyclic quadrilateral)
AED = 47 (alternate angles)
ADB = 47 (alternate segment theorem)

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AED = 47 (alternate angle)
BCD = 124 (cyclic theorum)
ADB = 47 (alternate segment)

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Angle ABC= 112 ( Cyclic Quadrilateral theorem)
Angle AOC= 136 (Angle made at the centre by an arc is twice the angle made at the circumference by the same arc)

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2) <AED=47
Alternate angles are equal
<BCD=124
<ADB=77
Alternate segment theorem.

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ABC= D+B will make 180
so, 68+B=180
180-68= 112 ABC= 112
the opposite angles of the cyclic quadrilateral are supplementary (theorem 6)

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AOC=136 ( the angle at the center is twice)
ABC= 112 ( the cyclic quadrilateral)

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Cyclic quadrilateral : 112 (180-68)

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Angle ABC is 112 using the theorem of cyclic quadrilateral

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Munahil-AOC=68*2=136(the angle in the center is twice of that at the circumference)

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Angle AOC is 68*2 = 136 as theorem 1 which means the angle at the center is twice the angle subtended at the circumference, applies here

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ABC=112 (cyclic quad)
AOC=136 (angle at center)

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Angle abc = 180 -68 = 112 (cyclic quadrilateral)

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AOC= 68X2 = 136
Theorem 1

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Munahil-180-68=112(cyclic quadrilateral, opposite angles add up to 180)

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opposite angles of a cyclic quadrilateral are supplimentary
Q3.a. PQS= 41
PTS= 75
Amnah

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ABC = 180-68 = 112
AOC = 136

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ABC = 112
AOC = 136

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Angle ABC = 112
the angles at the cyclic quadilateral are supplimentary.
AOC=136 the angle at the center is twice the angle at he circumference. so 68*2=136

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68 x 2 = 136
(theorem 1) the angle at the center is twice the angle at the circumference
ramsha

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For Angle AOC, the answer is 136 if we follow Theorm 1