AC=x, BC=x√3, AB is a hypotenuse (look at the pic below)
x√3>x, consequently ∠B<∠A and we have to find cosB, which is (x√3)/AB.
<em>Let's find AB:</em>
AB=√(x²+(x√3)²)=√(x²+3x²)=√(4x²)=2x
<em>Thus, cosB is:</em>
cosB=(x√3)/(2x)=√3/2
x√3>x, consequently ∠B<∠A and we have to find cosB, which is (x√3)/AB.
<em>Let's find AB:</em>
AB=√(x²+(x√3)²)=√(x²+3x²)=√(4x²)=2x
<em>Thus, cosB is:</em>
cosB=(x√3)/(2x)=√3/2