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प्रश्न
In a Δ ABC,∠A= x°,∠B = (3x × 2°),∠C = y° and ∠C - ∠B = 9°. Find the there angles.
उत्तर
∠C - ∠B = 9°
∴ y° – (3x – 2)° = 9°
⇒ y° – 3x° + 2° = 9°
⇒ y° – 3x° = 7°
The sum of all the angles of a triangle is 180°, therefore
∠A + ∠B + ∠C = 180°
⇒ x° + (3x – 2)°+ y° = 180°
⇒ 4x°+ y° = 182°
Subtracting (i) from (ii), we have
7x° = 182° – 7° = 175°
⇒ x° = 25°
Now, substituting x° = 25° in (i), we have
y° = 3x° + 7° = 3 × 25° + 7° = 82°
Thus
∠A = x° = 25°
∠B = (3x – 2)° = 75° - 2° = 73°
∠C = y° = 82°
Hence, the angles are 25°, 73° and 82°.
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