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# two parallel chords of a circle which are on the same side of the centre subtend angles of 72 degrees and 144 degrees respectively at the centreprove that the perpendicular distance between the chords is half the radius of the circle

Himanshu
103 Points
5 years ago
Let the two chords be AB and CD of the circle with centre O, where AB>CD.
AOB + OAB + OBA = 180`
144` + OAB + OAB = 180`
OBA=OAB=18`

COD + OCD + ODC = 180`
72` + OCD + ODC = 180`
OCD = ODC = 54`
Let OX be the perpendicular to chords which cut AB at M and CD at N.
In triangle OAM, sin18` = OM / radius
In triangle OBN, sin54` = ON / radius
radius × sin54` = ON               ….ii)

Subtract i) and ii) ,
ON – OM = radius (sin18` – sin54`)
NM = radius / 2 (sin18` – sin54`= 2)
Hence, proved.