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 ,

a

sin‌A

=

b

sin‌B

=

c

sin‌C

=k
⇒sin‌A=

a

k

,sin‌B=

b

k

,sin‌C=ck
Given ,sin‌A+sin‌B=sin
⇒A+B=C
⇒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,

a

sin‌A

=

b

sin‌B

=

c

sin‌C

=k

a

sin‌30

=

b

sin‌60

=

c

sin‌90

=k

a

1 2

=

b

√3 2

=

c

1

=k

a

1

=

b

√3

=

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:√3:2.