Question : O is the centre of a circle that passes through P,Q, R and S, as shown in the figure. SR is produced to X. If ∠QRX=133°, find x.
Doubt by Rishabh
Solution :
PQRS is a cyclic Quadrilateral
SRX is a straight light.
SRX is a straight light.
∠SRQ+133°=180° (Linear Pair)
∠SRQ=180°-133°
∠SRQ=47°
∠SRQ=180°-133°
∠SRQ=47°
Also,
∠SRQ+(4x+13)°=180° (Sum of opposite angles of a cyclic Quadrilateral is supplementary)
∠SRQ+(4x+13)°=180° (Sum of opposite angles of a cyclic Quadrilateral is supplementary)
47°+(4x+13)°=180°
(4x+13)°=180°-47°
(4x+13)°=133°
4x°+13°=133°
(4x+13)°=180°-47°
(4x+13)°=133°
4x°+13°=133°
4x°=133°-13°
4x°=120°
x=120°/4
x=30°