In the given figure, PQ is a diameter of a circle and PQ is parallel to RS, also angle PQR=32°. Find. (i) angle QSR. (ii) angle SQR
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(i) L PRQ = 90° (Angle of semi-circle)
In PQR,
RPQ + PRQ + PQR = 180°
RPQ + 90° + 32° = 180°
RPQ = 58°
Now,
RPQ + QSR = 180° (Opp. angles of cyclic quad.)
58° + QSR = 180°
QSR = 180° — 58° = 122°.
(ii) SRQ = PQR (Alternate angles as PQ II RS) = 32°
In RSQ.
SRQ + QSR + SQR = 180°
32°+ 122° + SQR = 180°
SQR= 26°
In PQR,
RPQ + PRQ + PQR = 180°
RPQ + 90° + 32° = 180°
RPQ = 58°
Now,
RPQ + QSR = 180° (Opp. angles of cyclic quad.)
58° + QSR = 180°
QSR = 180° — 58° = 122°.
(ii) SRQ = PQR (Alternate angles as PQ II RS) = 32°
In RSQ.
SRQ + QSR + SQR = 180°
32°+ 122° + SQR = 180°
SQR= 26°
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