# ISI2017-DCG-14

49 views

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. $1:3:2$

recategorized

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.

edited by

## Related questions

1
96 views
The area of the shaded region in the following figure (all the arcs are circular) is $\pi$ $2 \pi$ $3 \pi$ $\frac{9}{8} \pi$
If $a,b,c$ are the sides of $\Delta ABC$, then $\tan \frac{B-C}{2} \tan \frac{A}{2}$ is equal to $\frac{b+c}{b-c}$ $\frac{b-c}{b+c}$ $\frac{c-b}{c+b}$ none of these
The angle between the tangents drawn from the point $(-1, 7)$ to the circle $x^2+y^2=25$ is $\tan^{-1} (\frac{1}{2})$ $\tan^{-1} (\frac{2}{3})$ $\frac{\pi}{2}$ $\frac{\pi}{3}$
If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes are $(3,2)$, then the equation of the straight line is $2x+3y=12$ $3x+2y=0$ $2x+3y=0$ $3x+2y=12$