22 P、Q 和 R 是水平地面上的三个点，使得 RPQ 构成一个直角三角形，其中 PQ 和 RQ 的长度分别为 135 m 和 205 m。已知 P 处有一座高度为 50 m 的建筑物，且 Q 相对于 P 的方位角为 $119^\circ$。

1. Find the missing ground length $PR$ in the right-angled triangle $RPQ$.

In $\triangle RPQ$, right-angled at $P$,
$PR^2+PQ^2=RQ^2$
$PR^2=205^2-135^2=23800$
$PR=\sqrt{23800}\text{ m}$

2. Find the shortest horizontal distance from $P$ to the path $RQ$.

The largest angle of depression occurs at the point on $RQ$ closest to $P$.

Let this shortest distance be $d$ m. Using the area of $\triangle RPQ$,
$\frac12(PR)(PQ)=\frac12(RQ)(d)$
$d=\frac{PR\times PQ}{RQ}$
$d=\frac{\sqrt{23800}\times135}{205}$
$d\approx101.6\text{ m}$

3. Use trigonometry to calculate the largest angle of depression.

Let the largest angle of depression be $\theta$.

The building height is $50$ m and the horizontal distance is $d\approx101.6$ m.
$\tan\theta=\frac{50}{101.6}$
$\theta=\tan^{-1}\left(\frac{50}{101.6}\right)$
$\theta\approx26.2^\circ$