Question

In the figure given below, O is the centre of
the circle ZCED = 90º and ZABE
90= 120°.
Find ZBEC and ZBOC given that AC and
FD are parallel lines.​

Answer 1

Answer:

given that

\begin{gathered}\angle AOB = 90\\\angle BOC = 120\end{gathered}

∠AOB=90

∠BOC=120

let us find the angle AOC

\begin{gathered}\angle AOC = 360 - \angle AOB - \angle BOC \\\angle AOC = 360 - 90-120\\\angle AOC = 150\end{gathered}

∠AOC=360−∠AOB−∠BOC

∠AOC=360−90−120

∠AOC=150

Then

m(arc AC) = 150°

using inscribed angled theorem

\begin{gathered}m(\angle ABC ) = 0.5 \times m (arc AC) \\m(\angle ABC ) = 0.5 \times 150\\\angle ABC = 75\end{gathered}

m(∠ABC)=0.5×m(arcAC)

m(∠ABC)=0.5×150

∠ABC=75

hence ,

\angle ABC∠ABC = 75°