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प्रश्न
In Fig. AB is a diameter of a circle, with centre O. If ∠ABC = 70°, ∠CAD = 30° and ∠BAE = 60°, find ∠BAC, ∠ACD and ∠AВЕ.

योग
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उत्तर
Given
- AB is a diameter → ∠ACB = 90°
- ∠ABC = 70°
- ∠CAD = 30°
- ∠BAE = 60°
Must find ∠BAC, ∠ACD, and ∠ABE.
Find ∠BAC
In ΔABC, AB is a diameter ⇒ ∠ACB = 90°.
So,
∠BAC + ∠ABC + ∠ACB = 180°
∠BAC + 70° + 90° = 180°
∠BAC = 20°
Find ∠ABE
∠BAE = 60°
∠BAC = 20°
∠CAE = 60° − 20° = 40°
Now look at quadrilateral A–C–E–B on the circle.
Angles ∠CAE and ∠CBE stand on the same arc CE, so:
∠CBE = ∠CAE = 40°
In triangle ABC:
∠ABC = 70°
∠ABE + ∠CBE = 70°
∠ABE = 70° − 40° = 30°
Find ∠ACD
∠CBE = 40°
Angles ∠CBE and ∠ACD subtend the same arc CD.
∠ACD = ∠CBE = 40°
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