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
Answers
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°