sin ( A+B-C) =1/2 and cos (B+C-A)= 1/✓2 find the angles
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We know that in a triangle, sum of the angles = 180°
A+B+C = 180 → (1)
We know that,
So,
sin (A+B-C) = sin 30
A+B-C = 30 → (2)
And
cos (B+C-A) = cos 45
B+C-A = 45 → (3)
On solving equation (1) and (2), we get,
A+B+C-A-B+C = 180-30 = 150
2C = 150
C = 75°
Substituting C=75 in equation (2), we get,
A+B-75 = 30
A+B = 105 → (4)
Also, substituting in equation (3), we get,
B+75-A =45
A-B = 30 → (5)
Adding equations (4) and (5), we get,
2A = 135 → A = 67.5°
B = A-30 = 67.5 - 30 = 37.5°
Therefore, A=67.5°; B=37.5°; and C=75°
Explanation:
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