Question

in an isoceles triangle abc right angled at a the value of 2sin B cos C
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Answer 1

Answer:

ΔABC, 2sinCcosA=sinB [given]

⇒sin(C+A)+sin(C−A)=sinB

[∵2sinAcosB=sin(A+B)+sin(A−B)]

⇒sin(180

o

−B)+sin(C−A)=sinB

[∵∠A+∠B+∠C=180

o

]

⇒sinB+sin(C−A)=sinB

⇒sin(C−A)=0=sin0

o

⇒C−A=0

⇒C=A

Hence, ΔABC is an isosceles triangle.

Answer 2

Answer:

In the triangle ABCABC. A+B+C=πA+B+C=π. So, A=π−(B+C)A=π−(B+C). So, sinA=sin(π−(B+C))=sinπcos(B+C)−cosπsin(B+C)=sin(B+C)=sinBcosC+cosBsinCsin⁡A=sin⁡(π−(B+C))=sin⁡πcos⁡(B+C)−cos⁡πsin⁡(B+C)=sin⁡(B+C)=sin⁡Bcos⁡C+cos⁡Bsin⁡C.

We are given that this is equal to 2sinBcosC2sin⁡Bcos⁡C.

Hence 2sinBcosC=sinBcosC+cosBsinC⟹sinBcosC=cosBsinC2sin⁡Bcos⁡C=sin⁡Bcos⁡C+cos⁡Bsin⁡C⟹sin⁡Bcos⁡C=cos⁡Bsin⁡C,