Recent content by Tonyb24

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    Writing parallel vector equation

    if it passes through that point, then why not use that point as the support point? Then, what is a point on the Y axis that you can multiply to reach any y-axis point? So now, use your support point P and add to it a vector that can be multiplied by t where t E R, that will allow you to reach...
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    Solving for the third one should be straightforward.

    I know pi pi/2 pi/3 pi/4 and pi/6. I can do some fractions like 7pi/12 using ex.: sin(pi/4+pi/3)=sin(7pi/12)=Sin(pi/4)Cos(pi/3)+Sin(pi/3)Cos(pi/4) The don't use a calculator bit means exact answers as Mark said.
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    Solving for the third one should be straightforward.

    I squared 1+sqrt(3) as (1+sqrt(3))^2=(1+2sqrt(3)+3) =4+2sqrt(3) For context, this was on my test. Was wondering if someone could solve this from scratch. So plugging everything in b^2= (4+2sqrt(3))+4-(2(1+sqrt(3))(2)Cos(pi/3) add 4+4+2sqrt(3)=8+2sqrt(3), and finish the"-2acCos(pi/3)"...
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    Solving for the third one should be straightforward.

    Yes it is like that. Thanks.
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    Solving for the third one should be straightforward.

    Homework Statement Completely solve this triangle. No calculators please. A=? B=Pi/3 C=? a=(1+sqrt(3)) b=? c=2 Homework Equations Cosine law: b^2=a^2c^2-2ac(cos(B)) Sine law: Sin(A)/a=Sin(B)/b The Attempt at a Solution b^2=-6 You can plug in 1/2 in (cos(B)) right away. Other attemps, don't ask...