Question

in the adjoining figure, ABCD is a quadrilateral inscribed in a circle with center O. CD is produced to E. <ADE = 70° and <OBA =45°. Find;
(I) <OCA
(II) <BAC

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Answer 1

Answer:

<OCA = 20°

<BAC = 65°

ABCD is a Cyclic Quadrilateral

Given,

ABCD is a Quadrilateral inscribed in a circle with centre O.Also CD is a chord produced to E.

< ADE= 70°

<OBA =45°

<ADE + <ADC=180° (linear pair)

=> <ADC = 180°-70°

=110°

Again,

<D +< B= 180° ( opposite angle of a Cyclic Quadrilateral are supplementary)

=> 110°+<B =180°

=> <B= 180° - 110°

=> <B= 70°

Join OB, OA, OC,CA

=> <OBA +<OBC= 70°

=> 45° + < OBC=70°

=> < OBC = 25°

Also < AOC=140° (Angle subtended by an arc of a circle at the centre is double than the angle drawn subtended by it at any point on the remaining part of the circle)

Now in ∆ OAC

=> <OAC + <OCA + <AOC = 180°

=> <OAC + <OCA +140°= 180°

(OA= OC , radii of the same circle implies that angle opposite to equal sides are equal)

=> 2< OCA =40°

< OCA= 20°

similarly in ∆ OAB,

=> <OAB = <OBA =45° (•.• OA =OB, radii of same circle)

Now < BAC= <OAB + <OAC

< BAC =45°+20°

<BAC= 65°