Math, asked by deshpandeshilpa2237, 1 year ago

in a cyclic quadrilateral ABCD; AD=BC, angle BAC=30 degree and angle CBD=70 degree; find angle BCA.

Answers

Answered by Yogiraj1
17

ABCD is a cyclic quadrilateral whose diagonal intersect at E.


∠CBD = ∠CAD (Angles in the same segment are equal)

∠CAD = 70°

∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°

∠BCD + ∠BAD = 180° (Opposite angles of a cyclic quadrilateral)

∠BCD + 100° = 180°

∠BCD = 80°

In ΔABC,

AB = BC (Given)

∴ ∠BCA = ∠CAB (Angles opposite to equal sides of a triangle are equal)

⇒ ∠BCA = 30°

But ∠BCD = 80°.

⇒ ∠BCA + ∠ACD = 80°

30° + ∠ACD = 80°

⇒ ∠ACD = 50°

⇒ ∠ECD = 50°.
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