Question

in an acute angled triangle ABC ,if sin (A+B+C) =1/2 and cos (B+C-A)= 1/√2 find angle A ,B and C

Answer 1

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Answer 2

We have sin^ h ABC + − sin 2

1 = = 30º

or, ABC + − = 30º ...(1)

and cos^ h BCA + − cos

2

1 = = 45º

or, BCA + − = 45º ...(2)

Adding equation (1) and (2), we get

2B = 75º

or, B = 37 5º .

Now subtracting equation (2) from equation (1) we

get,

2^ h A C − =− 15º

or, A C − = 7 5º . ...(3)

Now ABC + + = 180º

ABC + + = 180º

A C + = 180 37 5 142 5 c c − . .º = ...(4)

Adding equation (3) and (4), we have

2A = 135º

or, A = 67 5º .

and, C = 75º

Hence, +A -67 .5° B- 37.5° C- 75º