1. ## angles problem

A man can row at 4 km/h, and run at 3 km/h. He needs to get from a point A, on the south bank of a stretch of still water, to point B on the north bank of the water. The direct distance from A to B is 5 km, and the water is 4 km wide. He starts rowing with an angle between North and the direction in which he rows. Find an expression for the time he will take to get from A to B, in terms of .

2. Hello, samtheman17!

A man can row at 4 km/h, and run at 3 km/h.
He needs to get from a point $A$, on the south bank of a stretch of still water,
to point $B$ on the north bank of the water.
The direct distance from $A$ to $B$ is 5 km, and the water is 4 km wide.
He starts rowing with an angle θ between North and the direction in which he rows.

Find an expression for the time he will take to get from $A$ to $B$, in terms of θ.
Code:
          3
C * - - - o B
|     *
4 |   * 5
| *
A o

The width of the water is: $AC = 4$ km.
The distance from $A$ to $B$ is: $AB = 5$ km.
Pythagorus tells us that: $CB = 3$ km.

He will start at $A$, row to $P$, then run to $B.$

Code:
          x   P   3-x
C * - - - * - - - - * B
|      /
|     /
|    / y
4 |   /
|  /
|θ/
|/
A *

Let $\theta \,=\, \angle CAP$
Let $x \,=\, CP$, then: $3-x \:=\:PB.$
Let $y \,=\,AP$

In right triangle $ACP\!:\;\;y^2 \:=\:x^2 + 4^2 \quad\Rightarrow\quad y \:=\:\sqrt{x^2+16}$

He rows $\sqrt{x^2+16}$ km at 4 km/hr.
. . This will take him: . $\frac{\sqrt{x^2+16}}{4}$ hours.

Then he runs $3-x$ km at 3 km/hr.
.This will take him: . $\frac{3-x}{3}\:=\:1-\tfrac{1}{3}x$ hours.

Hence, his total time is: . $T \;=\;\tfrac{1}{4}\sqrt{x^2+16} + 1 -\tfrac{1}{3}x$ .[1]

We want this answer in terms of θ.

In right triangle $ACP\!:\;\tan\theta \:=\:\frac{x}{4} \quad\Rightarrow\quad x \:=\:4\tan\theta$ .[2]

Note that: . $x^2+16 \;=\;(4\tan\theta)^2 + 16 \;=\;16\tan^2\!\theta + 16 \;=\;16(\tan^2\!\theta+1) \;=\;16\sec^2\!\theta$

. . Hence: . $\sqrt{x^2+16} \;=\;4\sec\theta$ .[3]

Subnstitute [2] and [3] into [1]: . $T \;=\;\tfrac{1}{4}(4\sec\theta) + 1 - \tfrac{1}{3}(4\tan\theta)$

Therefore: . $T \;=\;\sec\theta + 1 - \tfrac{4}{3}\tan\theta$

3. thank you so much!

i was trying to do something like that but kept messing something up as there are multiple steps....

it is so clear now, thanks again