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...
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.
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)"...
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...