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.
20^{o} - B.
30^{o} - C.
35^{o} - D.
40^{o} - E.
50^{o}
Correct Answer: Option E
Explanation
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°