ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC=70∘ and ∠BAC=30∘, then ∠BCD= ___. Further, if AB = BC, then ∠ECD = ___.
80∘,50∘
Since the angles in the same segment of a circle are equal,
∠CDB=∠BAC=30∘ ...(i)
Given, ∠DBC=70∘ .. .(ii)
Using angle sum property inΔBCD,
∠BCD+∠DBC+∠CDB=180∘.
Using (i) and (ii),
∠BCD+70∘+30∘=180∘.
∠BCD=180∘−100∘=80∘.....(iii)
In ΔABC, AB=BC
Since angles opposite to equal sides of a triangle are equal,
∴∠BCA=∠BAC=30∘ ...(iv)
We have ∠BCD=80∘
⇒∠BCA+∠ECD=80∘
i.e., 30∘+∠ECD=80∘
⇒∠ECD=80∘−30∘=50∘