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प्रश्न
In the given figure AC is the diameter of the circle with centre O. CD is parallel to BE.
∠AOB = 80° and ∠ACE = 20°.
Calculate
- ∠BEC
- ∠BCD
- ∠CED
उत्तर
a. ∠BOC = 180° – 80° = 100°
`\implies` ∠BEC = `1/2 xx 100^circ` = 50° ...(∠ at centre is twice the ∠ in remaining segment)
b. ∠BCD = ∠BCA + ∠ACE + ∠ECD
= 40° + 20° + 50°
= 110°
c. ∠CED = 180° – 110° – 50° = 20°
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