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Sin (a+b-c)=1⁄2 cos (a-b+c)=1 find the value of A, B and C?
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Sin (a+b-c)=1⁄2 cos (a-b+c)=1 find the value of A, B and C?
Understanding the Given Equations
The problem presents two trigonometric equations:
- Sin(a + b - c) = 1/2
- Cos(a - b + c) = 1
We need to find the values of angles A, B, and C.
Analyzing the First Equation
- The equation Sin(a + b - c) = 1/2 implies that:
- a + b - c = 30 degrees or 150 degrees (as Sin 30° = 1/2 and Sin 150° = 1/2).
Analyzing the Second Equation
- The equation Cos(a - b + c) = 1 indicates:
- a - b + c = 0 degrees (as Cos 0° = 1).
Setting Up the System of Equations
From the two equations, we can set up the following:
1. a + b - c = 30 or 150
2. a - b + c = 0
Solving the System
We will consider both cases for the first equation.
Case 1: a + b - c = 30
- Rearranging gives us c = a + b - 30.
- Substituting c in the second equation leads to a - b + (a + b - 30) = 0.
- Simplifying results in 2a - 30 = 0, hence a = 15 degrees.
- Then c = 15 + b - 30, which simplifies to c = b - 15.
Case 2: a + b - c = 150
- Similarly, we derive c = a + b - 150.
- Substituting in the second equation gives a - b + (a + b - 150) = 0.
- This leads to 2a - 150 = 0, hence a = 75 degrees.
- Thus, c = 75 + b - 150 simplifies to c = b - 75.
Conclusion
To summarize, we have two possible sets of solutions for (A, B, C):
- From Case 1: A = 15°, C = B - 15°
- From Case 2: A = 75°, C = B - 75°
You can choose values for B to derive specific solutions for (A, B, C).
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Sin (a+b-c)=1⁄2 cos (a-b+c)=1 find the value of A, B and C?
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