Question

please solve the problem

Answer 1

Answer:

ΔABC is isosceles.

Step-by-step explanation:

Given, A + B + C = π

⇒ C + A = π - B.

Now,

⇒ cosA = sinB/2sinC

⇒ 2sinCcosA = sin B

We know that 2sinacosb = sin(a + b) + sin(a - b).

⇒ sin(C + A) + sin(C - A) = sin B

⇒ sin(π - B) + sin(C - A) = sin B

⇒ sin B + sin(C - A) = sin B

⇒ sin(C - A) = 0

⇒ sin(C - A) = sin 0

⇒ C - A = 0

⇒ C = A

∴ΔABC is isosceles.

Hope this helps!

Answer 2

Step-by-step explanation:

By sine law,

sinBsinC=bc

By cosine law,

cosA=c2+b2−a22cb

Putting these in the given expression :

c^2+b^2−a22cb=b2c

c^2+b^2−a^2=b^2

c=a

Therefore,Δ ABC is isosceles.