Question

Plz solve it fast........​

Answer 1

Answer:

θ=0°

Step-by-step explanation:

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Sinθ = √3 (1-Cosθ)

Sinθ = √3 - √3 Cosθ

Sinθ + √3 Cosθ = √3

By dividing on both sides by $\sqrt{ 1^{2}+\sqrt{3} ^{2}   }  =2$ , we get,

(Sinθ)/2 + (√3 Cosθ)/2 = √3/2

Cos60*Sinθ + Sin60*Cosθ = Sin 60   [Cos60 = 1/2 and Sin60 = √3/2 ]

Sin(60+θ) = Sin60        [ Using Sin(A+B) =SinA*CosB+ CosA*SinB ]

60+θ= 60

θ= 0°

Answer 2

Answer:

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