Thread: Trigonometry cosine and sine rule

Number 11 in the attached document.

I don't know how to do it .. I can't figure out what i'm supposed to do. Help please

So you only want people who have Microsoft Word to respond?

Yes son.

Hi,
A=45°
B=60°
=> C=75°
You have to use for the both questions the sin-Formula:

i) sinA/a =sinB/b => sin45°/a=sin60°/b => (sin45°)^2/a^2=(sin60°)/b^2
you have to introduce the values for sin45° and sin60° and to transform the expression
ii) The solution is analog
to use: sin75°= cos15°; cos(2x)=2*cos(x))^2 -1 =>cos(x) =root((1+ cos2x)/2) => cos15°=root((2+root(3))/2) = 1/2*root(2*(1+ root(3))^2)= (1+root(3))/root(2),
=> sin75°= (1+root(3))/root(2)