Question

if a triangle ABC,if (a+b+c)(b+c-a)=3ABC then angel A equal to​

Answer 1

Answer:

Given:

(a+b+c)(b+c−a)=3bc

(b+c)

2

−a

2

=3bc

b

2

+c

2

−a

2

+2bc=3bc

2bc

b

2

+c

2

−a

2

=

2

1

According to the Law of Cosines:

cosA=

2bc

b

2

+c

2

−a

2

cosA=

2

1

∠A=

3

π

=60

o

Hence, option C is correct

Step-by-step explanation:

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