Circle Theorem Theorems Alternate Segment Opposite Angles Cyclic Quadrilateral Triangle Isosceles

[sunitakeny]

Tha kyou v.much for helping me through your video, s

[Grizzly]

Alternative method:
After using the alternate segment theorem to show ∠QSP = 41°, and declaring ∠SQR = x
you could then go on to show that due to SR = RQ, ∆QRS is isosceles, and thus ∠RSQ = ∠SQR = x
PQRS can be identified as a cyclic quadrilateral, so ∠PSR + ∠RQP = 180°
∴ 41 + x +57 + x = 180
∴ 2x + 98 = 180
Solving for x gives x = 41°

[pooransingh]

Circle theorem concepts need to be explored more fully at the high school level. Much learning exercises students.

[adgfx]

X=(180-98)/2=82/2=41 deg.oposit angles of cyclic quad.is supplimentary.

[Pryszczyk]

Alternative method:
Figure SRQP = isosceles trapeze
SR = RQ → ∠SPR = ∠RPQ
SR = RQ → ∠RSQ = ∠RQS = x
∠SRQ = ∠PQR = 57+x
SQR + 2x = 180
isosceles triangle → ∠SQR +2x = 180 = 57 +x +2x
180 = 57 +x +2x → 3x=123 → x= 41