Question

In the adjoining figure, ABCD is a cyclic quadrliateral in which $\angle BCD={100}^{\circ }and\angle ABD={50}^{\circ }.\text{Find}\angle ADB.$

Answer 1

ANSWER:
Given: ABCD is a cyclic quadrilateral.
∴ ∠ DAB + ∠ DCB = 180 ° ( Opposite angles of a cyclic quadrilateral are supplementary)
⇒ ∠ DAB + 100 ° = 180 °
⇒ ∠ DAB = (180 ° – 100 ° ) = 80 °
Now, in Δ ABD, we have:
⇒ ∠ DAB + ∠ ABD + ∠ ADB = 180 °
⇒ 80 ° + 50 ° + ∠ ADB = 180 °
⇒ ∠ ADB = (180 ° – 130 ° ) = 50 °
Hence , ∠ ADB = 50 °