UPSC NDA Math Model Paper 2

Let us consider Δ ABC with A, B and C as its angles and a, b and c as its sides.
Given: Angles are in AP i.e A, B and C are in AP.
⇒ 2 × B = A + C
As we know that, A + B + C =180°
⇒ 3 × B = 180° ----(∵ 2 × B = A + C)
⇒ B = 60°
As we know that, if A, B and C are angles and a, b and c are the sides of a Δ ABC, then:

a

sin‌A

=

b

sin‌B

=

c

sin‌C

⇒

b

c

=

sin‌B

sin‌C

As it is given that b:c=√3:√2 and sin‌C=sin‌60°=√3∕2.
⇒

√3

√2

=

√3

2

‌sin‌C⇒sin‌C=

1

√2

⇒C=45°
∵A+B+C=180° ⇒A+60°+45°=180°
⇒A=75°.