Question

If A+B-C= 180° and sin²A+sin²B-sin²C=K sin A sin B cos C, then the value of K is​

Answer 1

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Answer:

the value of k is 2

Step-by-step explanation:

sin^2A = 1-cos(2A)/2

same as for sin^2B and sin^2C

now substitute this value in the equation

you get

sin^2A+sin^2B+sin^2C=

sin^2A as,

sin^2A = (1-cos(2A))/2

Therefore,

LHS =

= (1-cos(2A))/2+(1-cos(2B))/2+(1-cos(2C))/2

=3/2 - (cos(2A)+cos(2B)+cos(2C))

substitute it with the equation give you get answer

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