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Chords AC and BD of a circle with centre O intersect at right angles at E. If ∠OAB = 25°, then the value of ∠EBC is
Answer & Solution
Correct Answer:
Option
B
According to question
Given:
∠OAB = 25°
OA = OB = r
∴ ∠OAB = ∠OBA = 25°
∴ ∠AOB = 180° - 25° - 25°
∠AOB = 130°
∴ ∠ACB = $$\frac{1}{2}$$∠AOB = $$\frac{{{{130}^ \circ }}}{2}$$ = 65°
In right angle ΔBEC
∠BEC + ∠EBC + ∠ECB = 180°
∠EBC = 180° - 65° - 90°
∠EBC = 25°
Given:
∠OAB = 25°
OA = OB = r
∴ ∠OAB = ∠OBA = 25°
∴ ∠AOB = 180° - 25° - 25°
∠AOB = 130°
∴ ∠ACB = $$\frac{1}{2}$$∠AOB = $$\frac{{{{130}^ \circ }}}{2}$$ = 65°
In right angle ΔBEC
∠BEC + ∠EBC + ∠ECB = 180°
∠EBC = 180° - 65° - 90°
∠EBC = 25°
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