Question

please answer 27th question

Answer 1

In fig.
ang DEC = 90°. (angle made on semicircle)
then ang CDE = 50°(bt angle sum prop of ∆)

now ang COP =AOD =70° (V.O.A)

and ang OPC = 70 ( by A.S.P. of∆)

ang BPE = OPC (V.O.A)
NOW in∆BPE
ang PBE= 20°(BY A.S.P.of ∆) =OBE


thus required
ang CDE = 50°
angOBE =20°

Answer 2

ur answer is in the attachment