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.
A.
20o
B.
30o
C.
35o
D.
40o
E.
50o
Correct answer is E
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°