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# Solving compound angle eqns watch

1. 3SinA= Cos(A+60°)
Ok i got the answer by expanding the rhs, replacing the known trig values by their fractions and dividing by CosA, which gives 3tanA + (√3/2)tanA= 1/4
Solving this i got 7.36, 187.36

However my 1st attempt at this question was trying to square both sides as below
3SinA= CosACos60°-SinASin60°
9(SinA)^2 = ((CosA)^2)/4 - (3(SinA)^2)/4
Replacing Cos^2 A by 1 - Sin^2 A

9Sin^2A= 1/4 - Sin^2A/4 - 3Sin^2/4
10Sin^2A= 1/4
A= Sin'(√(1÷40))
But this does not give me the answer above, anybody can spot a mistake?
2. (Original post by Bilbao)
3SinA= Cos(A+60°)
Ok i got the answer by expanding the rhs, replacing the known trig values by their fractions and dividing by CosA, which gives 3tanA + (√3/2)tanA= 1/4
Solving this i got 7.36, 187.36

However my 1st attempt at this question was trying to square both sides as below
3SinA= CosACos60°-SinASin60°
9(SinA)^2 = ((CosA)^2)/4 - (3(SinA)^2)/4
Replacing Cos^2 A by 1 - Sin^2 A

9Sin^2A= 1/4 - Sin^2A/4 - 3Sin^2/4
10Sin^2A= 1/4
A= Sin'(√(1÷40))
But this does not give me the answer above, anybody can spot a mistake?
On the RHS, (a - b)^2 = a^2 - 2ab + b^2 not a^2 - b^2.
3. (Original post by old_engineer)
On the RHS, (a - b)^2 = a^2 - 2ab + b^2 not a^2 - b^2.
Oh yea thanks

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