图中显示了一个花园中水平地面上的四个点 P、Q、R 和 S。 P 在 Q 的正南方。从 P 和 Q 观测 S 的方位角分别为 068° 和 126°。 PS = 460 米，QR = 562 米。 已知从 Q 观测 R 的方位角为 098°。

1. Find the relevant angles in triangle PQS from the given bearings.

Since P is due south of Q, Q is due north of P.

At P, the angle between $PQ$ and $PS$ is
$\angle QPS = 68^\circ.$

At Q, $QP$ is due south, so its bearing is $180^\circ$. Since the bearing of S from Q is $126^\circ$,
$\angle PQS = 180^\circ - 126^\circ = 54^\circ.$

2. Apply the sine rule to calculate QS.

In $\triangle PQS$, side $PS$ is opposite $\angle PQS = 54^\circ$, and side $QS$ is opposite $\angle QPS = 68^\circ$.

Using the sine rule,
$\frac{QS}{\sin 68^\circ}=\frac{460}{\sin 54^\circ}.$

$QS=\frac{460\sin 68^\circ}{\sin 54^\circ}=527.4\ldots$

So, to 3 significant figures,
$QS=527\text{ m}.$

1. Find the included angle $\angle SQR$ using the bearings from Q.

From Q, the bearing of R is $098^\circ$ and the bearing of S is $126^\circ$.

$\angle SQR = 126^\circ - 98^\circ = 28^\circ$

2. Apply the cosine rule to find $RS$.

Using $QS = 538$ m from part (a), $QR = 562$ m and $\angle SQR = 28^\circ$,

$RS^2 = 538^2 + 562^2 - 2(538)(562)\cos 28^\circ$

$RS^2 = 71358.\ldots$

$RS = \sqrt{71358.\ldots} = 267.\ldots$

1. Identify the angle between Joseph's path from Q towards P and the line QR.

Let Joseph be at $X$ on $PQ$ when he is west of $R$.

Since $P$ is due south of $Q$, $QX$ is due south.

The bearing of $R$ from $Q$ is $098^\circ$, so
$\angle XQR=180^\circ-98^\circ=82^\circ.$

2. Use trigonometry to find the distance from Q to Joseph's position.

In right-angled triangle $QXR$, $QR=562$ m and $QX$ is adjacent to $82^\circ$.

$\cos 82^\circ=\frac{QX}{562}$

$QX=562\cos82^\circ=78.2\text{ m}$