दिलेल्या आकृतीमध्ये, O हे वर्तुळाचे केंद्र आहे आणि ∠AOC = 140°. ∠ABC शोधा.

F1 SSC Arbaz 18-05-2023 Himanshu D2

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SSC CGL 2022 Tier-I Official Paper (Held On : 07 Dec 2022 Shift 4)
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  1. 95°
  2. 110°
  3. 120°
  4. 103°

Answer (Detailed Solution Below)

Option 2 : 110°
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दिलेले:

O हे वर्तुळाचे केंद्र आहे

∠AOC =140°

गणना:

O हे वर्तुळाचे केंद्र आहे

उर्वरित चाप मध्ये D बिंदू घ्या

F1 SSC Arbaz 18-05-2023 Himanshu D3

∠AOC = 140°

∠AOC = 2∠ADC (मध्यभागी जीवा द्वारे तयार केलेला कोन = समान चाप खंडात दोनदा कोन तयार होतो)

⇒ 140° =  2∠ADC

⇒ ∠ADC  = 70°

ABCD हा चक्रीय चौकोन असेल

चक्रीय चौकोनाच्या विरुद्ध कोनांची बेरीज = 180°

⇒ ∠ABC + ∠ADC= 180°

⇒ 70° + ∠ABC = 180°

⇒ ∠ABC = 110°

∴ पर्याय 2 हे योग्य उत्तर आहे.

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