WAEC - Mathematics (1990 - No. 36)

In the diagram above, PQ is a tangent at T to the circle ABT. ABC is a straight line and TC bisects ∠BTO. Find x.

20^o

30^o

35^o

40^o

50^o

From the figure < TAB = < BTQ = 40° (alternate segment)

\(\therefore\)< ATB = 180° - (70° + 40°) = 70° (angle on a straight line)

< BTC = \(\frac{40°}{2} = \frac{< BTQ}{2}\)

\(\therefore < BTQ = 40°\)

x° = 180° - (40° + 70° + 20°)

= 50°