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secondary 4 | A Maths
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need help with part d
Here, a = 12 and b = 5.
R = sqrt (12² + 5²)
R = 13
α = inverse tangent (5/12)
α = 22.61986495°
12 cos θ + 5 sin θ = 13 cos (θ - 22.6°)
For our expression in part (d), the value of the fraction is maximised when the denominator is minimised to a positive value which is close to zero.
Denominator = 15 + 13 cos (θ - 22.6°)
Since -1 ≤ cos (θ - 22.6°) ≤ 1, the smallest possible positive value of the denominator
= 15 + 13 (-1)
= 2
So, the greatest possible value of the fraction is 1/2.
This occurs when cos (θ - 22.6°) = -1.
θ - 22.6° = 180° (one value only)
θ = 202.6°
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