two parallel chords of a circle which are on the

Question

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 · Accepted Answer

Please draw a figure while reading the following explanation.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 / radiusradius × sin18`= OM….i)In triangle OBN, sin54` = ON / radiusradius × sin54` = ON….ii) Subtract i) and ii) ,ON – OM = radius (sin18` – sin54`)NM = radius / 2 (sin18` – sin54`= 2)Hence, proved.

Trigonometry> two parallel chords of a circle which are...

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 centre
prove that the perpendicular distance between the chords is half the radius of the circle

neel khare , 8 Years ago

Himanshu

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Trigonometry> two parallel chords of a circle which are...

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

neel khare , 8 Years ago

Himanshu

Provide a better Answer & Earn Cool Goodies

LIVE ONLINE CLASSES

Ask a Doubt

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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 centre
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Find the general solution of equation tan3∂=cot2∂
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