In the diagram, $\triangle P Q R$ is right-angled at $P$ and $\angle P R Q=\theta$. A circle with centre $P$ is drawn passing through $Q$. The circle intersects $P R$ at $S$ and $Q R$ at $T$. If $Q T=8$ and $T R=10$, determine the value of $\cos \theta$.

![](https://cdn.mathpix.com/cropped/2024_04_30_3011c729c030a190170fg-3.jpg?height=420&width=588&top_left_y=1357&top_left_x=823)
Solution 1

Let $S R=x$ and the radius of the circle be $r$.

Extend $R P$ to meet the circle again at $U$.

Then $P S=P Q=P U=r$, since each is a radius.

![](https://cdn.mathpix.com/cropped/2024_04_30_75c1cf70fbe3735bfd92g-12.jpg?height=412&width=594&top_left_y=409&top_left_x=820)

By the Pythagorean Theorem in $\triangle P Q R$, we have $P Q^{2}+P R^{2}=Q R^{2}$ or $r^{2}+(r+x)^{2}=18^{2}$.

By the Secant-Secant Theorem, $R S \cdot R U=R T \cdot R Q$ or $x(x+2 r)=10(18)$.

From the first equation, $r^{2}+r^{2}+2 x r+x^{2}=324$.

From the second equation, $x^{2}+2 x r=180$.

Substituting, we obtain $2 r^{2}+180=324$ or $r^{2}=72$ and so $r=\sqrt{72}=6 \sqrt{2}$ since $r>0$.

Since $\triangle P Q R$ is right-angled, then $\theta=\angle P R Q$ is acute, which means that $\cos \theta>0$.

Also,

$$
\cos ^{2} \theta=\left(\frac{P R}{R Q}\right)^{2}=\frac{R Q^{2}-P Q^{2}}{R Q^{2}}=\frac{324-72}{324}=\frac{252}{324}=\frac{7}{9}
$$

Therefore, $\cos \theta=\sqrt{\frac{7}{9}}=\frac{\sqrt{7}}{3}$.

## Solution 2

Let $S R=x$ and the radius of the circle be $r$.

Join $P T$. Then $P Q=P T=r$.

Since $Q T=8$ and $T R=10$, then $Q R=18$.

By the Pythagorean Theorem, $P R=\sqrt{Q R^{2}-P Q^{2}}=\sqrt{18^{2}-r^{2}}$.

Therefore, $\cos \theta=\frac{P R}{Q R}=\frac{\sqrt{18^{2}-r^{2}}}{18}$.

By the cosine law in $\triangle P T R$,

$$
\begin{aligned}
P T^{2} & =P R^{2}+T R^{2}-2(P R)(T R) \cos \theta \\
r^{2} & =\left(18^{2}-r^{2}\right)+10^{2}-2 \sqrt{18^{2}-r^{2}}(10) \cdot \frac{\sqrt{18^{2}-r^{2}}}{18} \\
r^{2} & =324-r^{2}+100-\frac{10}{9}\left(324-r^{2}\right) \\
r^{2} & =424-r^{2}-360+\frac{10}{9} r^{2} \\
\frac{8}{9} r^{2} & =64 \\
r^{2} & =72
\end{aligned}
$$

Therefore, $\cos \theta=\frac{\sqrt{324-r^{2}}}{18}=\frac{\sqrt{252}}{18}=\frac{\sqrt{7}}{3}$.