UPSC NDA I and II 2020 Math Solved Paper

(1) ΔABC is a right angled triangle.
sin‌B=

AC

BC

=

1

3

∴AC=1,BC=3
Now,
AB=√(BC)2−(AC)2 =√(3)2−(1)2=2√2
Now, sin‌C=

AB

BC

=

2√2

3

∴cosec‌C=

3

2√2

Hence, (1) is not correct.
(2) b.cos‌B=c.cos‌C
b.

(a2+c2−b2)

2ac

=c.

(a2+b2−c2)

2ab

b2(a2+c2−b2)=c2(a2+b2−c2)
a2b2−b4−a2c2+c4=0
a2(b2−c2)−(b4−c4)=0
(b2−c2)(a2−b2−c2)=0
Either b2+c2−a2=0 or (b2−c2)=0
When, b2+c2−a2=0
b2+c2=a2
Hence, ΔABC is a right angle triangle.
And when, b2−c2=0⇒b=c
Hence, ΔABC is an isosceles triangle.
From question ΔABC is not right angle triangle.
Hence, ΔABC must be an isosceles triangle.