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In an acute angled triangle ABC, if sin (2A + B - C) = 1 and tan (B + C - A) = √3. then find the values of A, B and C?
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In an acute angled triangle ABC, if sin (2A + B - C) = 1 and tan (B + ...
Given Information:
In an acute angled triangle ABC,
- sin (2A + B - C) = 1
- tan (B + C - A) = √3

Solution:

Using Trigonometric Identities:
We know that sin2x + cos2x = 1
Therefore, sin2A + cos2A = 1
=> 1 - cos2A + cos2A = 1
=> cos2A = 0
=> cosA = 0
=> A = 90°

Using tan(B + C - A) = √3:
tan(B + C - 90°) = √3
tan(B + C - 90°) = tan60°
B + C - 90° = 60°
B + C = 150°

Using sin(2A + B - C) = 1:
sin(2*90° + 150° - C) = 1
sin(180° + 150° - C) = 1
sin(330° - C) = 1
sin(30°) = 1
C = 30°

Final Values:
A = 90°
B = 60°
C = 30°
Therefore, the values of A, B, and C in the acute-angled triangle ABC are 90°, 60°, and 30° respectively.
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In an acute angled triangle ABC, if sin (2A + B - C) = 1 and tan (B + C - A) = √3. then find the values of A, B and C?
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