Q.If sin(45° + θ) = cos(2θ - 30°), then θ = ? (a) 25° (b) 45° (c) 15° (d) 35°
Answer: C
\(sin⁡(45°+θ)=cos⁡(2θ-30°)\)
\(=sin⁡[90°-(2θ-30°) ] \)
\(∴ 45°+θ=90°-(2θ-30°)\)
or, \(45°+θ=60°-2θ \)
or, \(θ+2θ=60°-45° \)
or, \(θ=15° \)
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