UPSC NDA Math Model Paper 3

Consider, a, b, c are the sides of triangle and A, B, C are the angles of the triangle. We know that $, \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ $\Rightarrow \sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = ck$ Given $, \sin A + \sin B = \sin$ $\Rightarrow A + B = C$ $\Rightarrow a + b = c$ This is not posible. Hence, there exists no triangle ABC for which sin A + sin B = sin C. (2) Given, the angles of a triangle are in the ratio 1 : 2 : 3. Consider the angles of a triangle are A = x, B = 2x and C = 3x. We know that, Sum of the angles of the triangle is 180° Hence, x + 2x + 3x = 180 6x = 180° x = 30º Angles are A = 30º, B = 60º and C = 90º We know that, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ $\frac{a}{\sin 30} = \frac{b}{\sin 60} = \frac{c}{\sin 90} = k$ $\frac{a}{\frac{1}{2}} = \frac{b}{\frac{\sqrt{3}}{2}} = \frac{c}{1} = k$ $\frac{a}{1} = \frac{b}{\sqrt{3}} = \frac{c}{2} = k$ Hence, the angles of a triangle are in the ratio 1 : 2 : 3, then its sides will be in the ratio $1 : \sqrt{3} : 2.$