Answer
(a) See explanations below.
(b) $12cm$
(c) vertical plane.
Work Step by Step
(a) Use the Law of Sines in triangle ABC, we have $\frac{BA}{sin60^\circ}=\frac{b}{sin\angle B}$ which gives $BA=b\frac{sin60^\circ}{sin\angle B}$.
Similarly in triangle ADC, we have $\frac{AD}{sin60^\circ}=\frac{b}{sin\angle D}$ which gives $AD=b\frac{sin60^\circ}{sin\angle D}$.
And in triangle BCD, we have $\frac{BD}{sin120^\circ}=\frac{a}{sin\angle D}$ which gives $BD=a\frac{sin120^\circ}{sin\angle D}$.
Use the relationship $BD=BA+AD$ and ${sin120^\circ}={sin60^\circ}$, we get $a\frac{sin60^\circ}{sin\angle D}=b\frac{sin60^\circ}{sin\angle B}+b\frac{sin60^\circ}{sin\angle D}$.
Cancel $sin60^\circ$ and multiply $sin\angle D$ on both sides, we get $a=b\frac{sin\angle D}{sin\angle B}+b$ which leads to $\frac{sin\angle D}{sin\angle B}=\frac{a-b}{b}$
Again in triangle BCD, we have $\frac{r}{sin\angle B}=\frac{a}{sin\angle D}$ which gives $r=a\frac{sin\angle B}{sin\angle D}=\frac{ab}{a-b}$.
(b) Given $a=4,b=3$, we have $r=\frac{4\times3}{4-3}=12cm$
(c) When $a=b$, $r$ will be $\infty$ which means that the common face would be a vertical plane.
