Hello, squeeze101!

A man can row at 8 km/h, and run at 15 km/h.

He needs to get from a point $\displaystyle A$, on the south bank of a stretch of still water,

to point $\displaystyle B$ on the north bank of the water.

The direct distance from $\displaystyle A$ to $\displaystyle B$ is 17 km, and the water is 8 km wide.

He starts rowing with an angle $\displaystyle \theta$ between North and the direction in which he rows.

Find an expression for the time he will take to get from $\displaystyle A$ to $\displaystyle B$, in terms of $\displaystyle \theta$. Code:

: - - - - - - - 15 - - - - - - - - :
C 8tanθ P 15-8tanθ B
* - - - - - - - - * - - - - - - - - * - -
| * *
| * *
8 | * * 17
| θ * *
| * *
A * - - - - - - - - - - - - - - - - - - - -

The distance directly across the water is: $\displaystyle AC = 8$

We are told that: $\displaystyle AB = 17$

From Pythagorus, we find that: $\displaystyle CB = 15$

He will row toward point $\displaystyle P$ at an angle of $\displaystyle \theta = \angle CAP$

In right triangle $\displaystyle PCA\!:\;\;AP = 8\sec\theta$

. . and: .$\displaystyle CP = 8\tan\theta \quad\Rightarrow\quad PB \:=\:15-8\tan\theta $

He will row $\displaystyle AP \:=\: 8\sec\theta$ km at 8 km/hr

. . This will take him: .$\displaystyle \frac{8\sec\theta}{8} \:=\:\sec\theta\,\text{ hours.}$

He will run $\displaystyle PB \:=\:15-8\tan\theta$ km at 15 km/hr.

. . This will take him: .$\displaystyle \frac{15-8\tan\theta}{15} \:=\:1 - \tfrac{8}{15}\tan\theta\,\text{ hours.}$

Therefore, his total time is: .$\displaystyle T \;=\;\sec\theta + 1 - \tfrac{8}{15}\tan\theta\,\text{ hours} $