In a $△ A B C, ∠ A = x, ∠ B = 3 x$ and $∠ C = y$ . If $3 y - 5 x = 30$ , then prove that the triangle is right angled triangle.

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Solution

$G i v e n : A = x, B = 3 x, C = y$
$∴ S u m o f △ l e = x + 3 x + y$
$180 = 4 x + y$
$⟹ 180 - 4 x = y$
$A l s o, 3 y - 5 x = 30^{\circ}$
Subtituting y value, $∴ 3 (180 - 4 x) - 5 x = 30$
$∴ x = 30, b y s o l v i n g$
$∴ y = 180 - 4 \times 30 = 60$
$∴ A = 30^{\circ}; B = 90^{\circ}; C = 60^{\circ}$

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Similar questions

In a $△ A B C, ∠ A = x^{\circ}, ∠ B = (3 x - 2)^{\circ}, ∠ C = y^{\circ}$ . If 3y – 5x = 30. Prove that the triangle is right angled.

∆ A B C, ∠ A = x °, ∠ B = (3 x) ° and ∠ C = y ° .

3 y - 5 x = 30,

Δ A B C, ∠ A = x^{\circ}, ∠ B = 3 x^{\circ}

∠ C = y^{\circ} .

5 x - 3 y + 30 = 0,

Δ A B C, ∠ A = x^{o}, ∠ B = 3 x^{o}

∠ C = y^{o}

3 y^{o} - 5 x^{o} = 30^{o}

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