24. In the diagram shown, $M$ is the mid-point of $P Q$. The line $P S$ bisects $\angle R P Q$ and intersects $R Q$ at $S$. The line $S T$ is parallel to $P R$ and intersects $P Q$ at $T$. The length of $P Q$ is 12 and the length of $M T$ is 1 . The angle $S Q T$ is $120^{\circ}$.
What is the length of $S Q$ ?

A 2
B 3
C 3.5
D 4
E 5
SOLUTION
B
Let $t$ be the length of $S Q$.
Let $\angle Q P S=x^{\circ}$.
Because $P Q$ has length 12 and $M$ is the midpoint of $P Q$, it follows that $M P$ has length 6 . We are given that $M T$ has length 1 . Therefore $P T$ has length 7 , and $T Q$ has length 5 .

Because $P S$ bisects $\angle R P Q$, we have $\angle R P S=$ $\angle Q P S=x^{\circ}$. Therefore $\angle R P Q=2 x^{\circ}$.
Because $S T$ is parallel to $P R$, we have $\angle S T Q=\angle R P Q=2 x^{\circ}$.
It follows from the External Angle Theorem [See Problem 24.1], applied to the triangle PTS, that $\angle P S T=x^{\circ}$.
Therefore the triangle $P S T$ is isosceles and hence $S T$ has the same length as $P T$, namely 7 . We now apply the Cosine Rule to the triangle STQ. This gives
$$
7^{2}=5^{2}+t^{2}-2\left(5 t \cos 120^{\circ}\right)
$$

Hence, as $\cos 120^{\circ}=-\frac{1}{2}$,
$$
49=25+t^{2}+5 t \text {. }
$$

Therefore $t^{2}+5 t-24=0$. Hence $(t-3)(t+8)=0$. Therefore $t$ is either 3 or -8 . Since $t$ corresponds to a length, $t>0$. We deduce that $t=3$.
For INVESTIGATION
24.1 The External Angle Theorem says that the external angle of a triangle is the sum of the two opposite angles.
In terms of the diagram it says that $\alpha=\beta+\gamma$.
Explain why the External Angle Theorem is true.
24.2 The solution above uses the fact that $\cos 120^{\circ}=-\frac{1}{2}$. Explain why this is correct.