ΔABC में, O लंबकेंद्र है और I अंतःकेंद्र है, यदि ∠BIC - ∠BOC = 90 है, तो ∠A ज्ञात कीजिए।

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  1. 120
  2. 140
  3. 90
  4. 180

Answer (Detailed Solution Below)

Option 1 : 120
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Detailed Solution

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दिया गया है:

ΔABC में, O लंबकेंद्र है और I अंतःकेंद्र है,

यदि ∠BIC - ∠BOC = 90 है,

प्रयुक्त सूत्र:

(1) ΔABC में, I अंतःकेंद्र है,

F1 Madhuri Railways 22.06.2022 D12

(1.1) ∠BIC = 90 + \(\frac{1}{2}\)∠A

(1.2) ∠AIC = 90 + \(\frac{1}{2}\)∠B

(1.3) ∠AIB = 90 + \(\frac{1}{2}\)∠C

(2) ΔABC में,लंबकेंद्र है,

F1 Madhuri Railways 22.06.2022 D13

(2.1) ∠BOC = 180∘ - ∠A

(2.2) ∠AOB = 180∘ - ∠C

(3.3) ∠AOC = 180∘ - ∠B

गणना:

प्रश्न के अनुसार, अभीष्ट आकृति है:

F1 Madhuri Railways 22.06.2022 D14

चूंकि हम जानते हैं,

∠BOC = 180∘ - ∠A     ----(1)

∠BIC = 90 + \(\frac{1}{2}\)∠A    ----(2)

अब, समीकरण (1) को (2) में से घटाने पर,

⇒ ∠BIC - ∠BOC = 90 + \(\frac{1}{2}\)∠A - (180∘ - ∠A )

⇒ 90 = 90 + \(\frac{1}{2}\)∠A - 180 + ∠A

⇒ 90 = \(\frac{3}{2}\)∠A - 90

⇒ 180 = \(\frac{3}{2}\)∠A

⇒ ∠A = 120

∴ अभीष्ट उत्तर 120 है।

Additional Information

(1) अंतःकेंद्र - यह त्रिभुज के तीनों कोणार्धों का प्रतिच्छेदन बिंदु होता है।

(1.1) कोणार्ध कोण को दो बराबर आधे में काटता है।

(2) लंबकेंद्र - यह शीर्षलंब से त्रिभुज के सम्मुख भुजा पर खींचे गए सभी तीन शीर्षलंबों का प्रतिच्छेदन बिंदु होता है।

(2.1) एक त्रिभुज का शीर्षलंब सम्मुख भुजा के लंबवत होता है।

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