Question

sin(2A+2B-2C) =1 and tan(B+C-A) =√3, find A, B and C​

Answer 1

Answer:

A = 60°, B = 52.5°, C = 67.5°

Step-by-step explanation:

sin(2A+2B-2C) =1

=> 2A+2B-2C = 90°

=> (A+B-C) = 45  ____(1)

tan(B+C-A) =√3

=> B+C-A = 60° ___(2)

Adding 1 and 2,

(A+B-C) + (B+C-A) = 45° + 60

2B = 105°

=> B = 52.5°

Now, substituting the value of B in (1)

A+52.5-C = 45

=> C-A = 7.5 ___(3)

We know that sum of all angles of triangle are 180°

A+B+C = 180°

=> A+52.5+C = 180

=> A+C = 127.5 ____(4)

Adding (3) and (4),

(C-A) + (A+C) = 7.5 + 127.5

=> 2C = 135

=> C = 67.5

Substituting the value of C in (4),

A+67.5 = 127.5

=> A = 127.5-67.5

=> A = 60°

Hope it helps

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