Hi, I've been having trouble with this question, and would really like some help.
What I've done so far is applied the cosine rule in the triangle PQR to find that $PR^2=a^2+c^2+2ac\cos\theta$.
What do I do from here? Thanks in advance.
EDIT: I know of the circle properties, in that opposite angles are supplementary for example, but how do I derive the LHS of the required solution?
On
Since the diameter $D$ of the circle is $QS=\sqrt{a^2+b^2}$, which means that the diameter of the circumscribed circle of $\triangle PSR$ is $\sqrt{a^2+b^2}$, using the law of sines for $\triangle PSR$, we have $$\frac{PR}{\sin\theta}=\sqrt{a^2+b^2}\Rightarrow PR=\sqrt{a^2+b^2}\sin\theta.$$
So, we have $$(PR^2=)\ \ \ (a^2+b^2)\sin^2\theta=a^2+c^2+2ac\cos\theta.$$
Hope the following construction helps.
The blue line is the diameter.
The green line is also a diameter.
The yellow-marked angle = $\theta$ (angle in the same segment)
The square root of the RHS of the to-be-shown = the length of the red line.
Consider the triangle formed by the black, red, green lines.