खालील आकृतीत 'x' चे मूल्य:

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HP JBT TET 2013 Official Paper
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  1. 60°
  2. 75°
  3. 80°
  4. 85°

Answer (Detailed Solution Below)

Option 1 : 60°
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HP JBT TET 2021 Official Paper
150 Qs. 150 Marks 150 Mins

Detailed Solution

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दिलेले आहे:

आकृती =

संकल्पना:

पूरक कोनांची बेरीज = 180°

त्रिकोणाच्या सर्व अंतर्गत कोनांची बेरीज = 180°

गणना:

⇒ ΔABC मध्ये

⇒ ∠ADB + ∠ADC = 180°

⇒ 108° + ∠ADC = 180°

⇒ ∠ADC = 180° - 108° = 72°

⇒ ΔADC मध्ये

⇒ ∠ADC + ∠ACD + ∠DAC = 180°

⇒ 72° + x° + 48° = 180°

⇒ x° = 180° - 48° - 72° = 180° - 120° = 60°

∴ आवश्यक उत्तर 60° असेल.

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