2017 PJC Promo Q9

In the triangle \(ABC\) as shown below, \(BC=3\), angle \(BAC=\frac{\pi }{3}+\theta \) radians and angle \(ACB=\frac{\pi }{2}\) radians.

Show that \(AC=\frac{3\left( 1-\sqrt{3}\tan \theta \right)}{\sqrt{3}+\tan \theta }\).[3]

Given that \(\theta \) is a sufficiently small angle, deduce that \(AC\approx \sqrt{3}+a\theta \), where \(a\) is a constant to be determined.[3]