In ∆ABC, if ∠A + ∠B = 108° and ∠B + ∠C = 130°, find ∠A, ∠B and ∠C.

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Solution

Let $∠ A + ∠ B = 108 ° and ∠ B + ∠ C = 130 °$ .
$\Rightarrow ∠ A + ∠ B + ∠ B + ∠ C = (108 + 130) ° \Rightarrow (∠ A + ∠ B + ∠ C) + ∠ B = 238 ° [∵ ∠ A + ∠ B + ∠ C = 180 °] \Rightarrow 180 ° + ∠ B = 238 ° \Rightarrow ∠ B = 58 °$

$∴ ∠ C = 130 ° - ∠ B$
$= (130 - 58) ° = 72 °$

$∴ ∠ A = 108 ° - ∠ B$
$= (108 - 58) ° = 50 °$

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Similar questions

$I n Δ A B C, i f ∠ A + ∠ B = 108^{\circ} a n d ∠ B + ∠ C = 130^{\circ}, f i n d ∠ A, ∠ B a n d ∠ C .$

In the given figure, $∠$ A= 130 degrees and $∠$ B= 40 degrees. If $△$ ABC is congruent to $△$ DEF, then find $∠$ F.

∠ A = 50^{\circ}

∠ B = 75^{\circ}

∠ C

∠ D .

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