Can someone find the values Θ when 0 < Θ < 180
Please this exercise was in Math Extend Awards 2011 and I cant find the solution...
Please include the general solution.
tan (θ+35°) = cot (θ-53°)
=>sin (θ+35°)/cos (θ+35°) = cos (θ-53°)/sin (θ-53°)
=> cos (θ+35°).cos (θ-53°) = sin (θ+35°).sin (θ-53°)
=> cos (θ+35°).cos (θ-53°) - sin (θ+35°).sin (θ-53°) = 0
=> cos (θ+35°+ θ-53°) = 0 [ as cos A.cos B - sin A.sin B = cos (A+B) ]
=> cos (2θ - 18°) = 0
=> 2θ - 18° = (2n + 1).90°, where n = 0, ±1, ±2, ±3....
=> θ = (2n + 1).45° + 9°, where n = 0, ±1, ±2, ±3....
Among all the values of θ, only 54° (for n =0) & 144° (for n = 1) lie in the interval (0°,180°)
Hence, 54° & 144° are the answers.
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tan (θ+35°) = cot (θ-53°)
=>sin (θ+35°)/cos (θ+35°) = cos (θ-53°)/sin (θ-53°)
=> cos (θ+35°).cos (θ-53°) = sin (θ+35°).sin (θ-53°)
=> cos (θ+35°).cos (θ-53°) - sin (θ+35°).sin (θ-53°) = 0
=> cos (θ+35°+ θ-53°) = 0 [ as cos A.cos B - sin A.sin B = cos (A+B) ]
=> cos (2θ - 18°) = 0
=> 2θ - 18° = (2n + 1).90°, where n = 0, ±1, ±2, ±3....
=> θ = (2n + 1).45° + 9°, where n = 0, ±1, ±2, ±3....
Among all the values of θ, only 54° (for n =0) & 144° (for n = 1) lie in the interval (0°,180°)
Hence, 54° & 144° are the answers.