Question

In the given figure, AC is a tangent to circle at point B. ∆EFD is an equilateral triangle and ∠CBD = 40°. Find:

1. ∠BFD
2. ∠FBD
3. ∠ABF

Answer 1

Given - In Diagram, ∆FED is an equilateral triangle, ∠CBD = 40°

To Find -

1. ∠BFD,
2. ∠FBD,
3. ∠ABF

(a)  ∠BFD = ∠CBD = 40°    ....[Alternate segment theorem]

(b) ∠FBD is opposite to ∠FED.

So, It is a cyclic Quadrilateral. So, ∠FBD and ∠FED will be Supplementary to each other.

∴ ∠FBD + ∠FED = 180°  ....[Cyclic Quadrilateral]

∠FBD + 60° = 180°  

∠FBD = 180° - 60°

∠FBD = 120°   

(c) AC is a line segment. 

So, ∠ABF = 180° - (120° + 40°)   ....[Line Segment]

∠ABF = 180° - 100°

∠ABF = 20°