In a triangle ABC, $b = \sqrt{3},$ c = 1 and $∠ A = 30^{0}$ , then the largest angle of the triangle is

60°

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135°

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90°

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120°

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Solution

The correct option is D
120°

Given that, $b = \sqrt{3}, c = 1 a n d ∠ A = 30^{0}$
$By cosine rule, cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
$\Rightarrow c o s 30^{0} = \frac{3 + 1 - a^{2}}{2 \sqrt{3}}$
$\Rightarrow \frac{\sqrt{3}}{2} = \frac{4 - a^{2}}{2 \sqrt{3}}$
$\Rightarrow 4 - a^{2} = 3$
$\Rightarrow a^{2} = 1$
$∴ ∠ A = ∠ C = 30^{0}$
$\Rightarrow ∠ B = 120^{0}$

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In a triangle ABC, b=√3, c = 1 and ∠A=300, then the largest angle of the triangle is

The correct option is D 120° Given that, b=√3,c=1 and ∠A=300 By cosine rule, cos A=b2+c2−a22bc ⇒cos 300=3+1−a22√3 ⇒ √32=4−a22√3 ⇒ 4−a2=3 ⇒ a2=1 ∴ ∠A=∠C=300 ⇒ ∠B=1200

In a triangle ABC, $b = \sqrt{3},$ c = 1 and $∠ A = 30^{0}$ , then the largest angle of the triangle is