\(a,BC=\dfrac{AB}{\cos B}=6:\dfrac{1}{2}=12\left(cm\right)\\ AC=\sqrt{BC^2-AB^2}=6\sqrt{3}\left(cm\right)\left(pytago\right)\\ b,P=\dfrac{\sin^2B}{\cos^2B}\cdot\left(\sin^2B+\cos^2B+\cos^2B-1\right)\\ P=\dfrac{\sin^2B}{\cos^2B}\left(1-1+\cos^2B\right)=\dfrac{\sin^2B}{\cos^2B}\cdot\cos^2B\\ P=\sin^2B=1-\cos^2B=1-\dfrac{1}{4}=\dfrac{3}{4}\)