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If $a,b,c$ are the sides of a triangle such that $a:b:c=1: \sqrt{3}:2$, then $A:B:C$ (where $A,B,C$ are the angles opposite to the sides of $a,b,c$ respectively) is

1. $3:2:1$
2. $3:1:2$
3. $1:2:3$
4. $1:3:2$

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Answer: $\text C$

Let $\mathrm{a = 1.x, \;b = \sqrt 3x, ~c= 2x}$

Now,

$\mathrm{a^2 + b^2 = x^2 + (\sqrt3x)^2 = x^2 + 3x^2 = 4x^2 = \bf{c^2}}$

$\Rightarrow$This is a Right Triangle.

Now, the angle opposite to the biggest side will be the right angle.

$\therefore \mathrm{\angle C \text{ is the right angle.}}$

Now,

$\mathrm {\sin A = \frac{a}{c} = \frac{x}{2x} = \frac{1}{2} \Rightarrow A = 30^0}$

Now,

$\mathrm{\angle C = 180^0 - 30^0 = 60^0}$

$\therefore \mathrm{A:B:C = 30:60:90 = 1:2:3}$

$\therefore \mathbf C$ is the correct option.

by Boss (13.4k points)
edited by