WAEC - Mathematics (1995 - No. 31)

In the diagram above; O is the centre of the circle and |BD| = |DC|. If ∠DCB = 35°, find ∠BAO.

20^o

25^o

30^o

35^o

40^o

< DBC = 35° (base angles of an isosceles triangle)

< CDB = 180° - (35° + 35°)

= 110°

< ADB = 70°; < ABD = 90°(< in a semi-circle)

\(\therefore\) < BAO = 180° - (70° + 90°)( sum of <s in a triangle)

= 20°