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Answer:
∠A = 20°
∠B = 40°
∠C = 120°
Step-by-step explanation:
we know that sum of all angles in a triangle is equal to 180°
therefore
∠A+∠B+∠C=180°
TYPE 1
∠A+∠B+3∠B=180° ←given in the question
∴ ∠A + 4∠B = 180° →⊂ 1 ⊃
TYPE 2
∠A+∠B+2(∠A+∠B) = 180° ←given in the question
∴ 3∠A+3∠B=180°
∠A+∠B=60° → ⊂ 2 ⊃
SUBRACTING ⊂2⊃ BY ⊂1⊃ WE GET
∠A + ∠B = 60°
∠A + 4∠B = 180°
-3∠B = -120°
∴ ∠B = 40°
∠A + 40° = 60° ← from ⊂2⊃
∴ ∠A = 20°
∠C = 3∠B ←←given in the question
∴ ∠C = 120°
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Answer:
∠A = 20°
∠B = 40°
∠C = 120°
Step-by-step explanation:
we know that sum of all angles in a triangle is equal to 180°
therefore
∠A+∠B+∠C=180°
TYPE 1
∠A+∠B+3∠B=180° ←given in the question
∴ ∠A + 4∠B = 180° →⊂ 1 ⊃
TYPE 2
∠A+∠B+2(∠A+∠B) = 180° ←given in the question
∴ 3∠A+3∠B=180°
∠A+∠B=60° → ⊂ 2 ⊃
SUBRACTING ⊂2⊃ BY ⊂1⊃ WE GET
∠A + ∠B = 60°
∠A + 4∠B = 180°
-3∠B = -120°
∴ ∠B = 40°
∠A + 40° = 60° ← from ⊂2⊃
∴ ∠A = 20°
∠C = 3∠B ←←given in the question
∴ ∠C = 120°
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