In a circle with centre 0, chord AB and diameter CD intersect each other at point E, inside the circle. If ∠AOD = 42° and ∠BOC = 104°, then what is the measure (in degrees) of ∠AED?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 73.
∠BAC = ∠EAC = ∠AOC/2 = 104/2 = 52°
∠AOC = 180 - 42 = 138
In triangle ACO
∠CAO + ∠ACO + ∠AOC = 180
∠CAO + ∠ACO + 138 = 180
∠CAO + ∠ACO = 42
OA and OC are radius
So, ∠CAO = ∠ACO = 21° = ∠ACE
In triangle ACE,
∠EAC + ∠ACE + ∠CEA = 180
52 + 21 + ∠CEA = 180
∠CEA = 180 - 73 = 107
So, ∠AED = 180 - 107 = 73