# Thread: In Terms of - Boating

1. ## In Terms of - Boating

A man can row at 24 km/h, and run at 7 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 25 km, and the water is 24 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.

I've been stuck on this question for quite sometime now.. would anybody kindly help me out?

Thank you!

2. Originally Posted by gearshifter
A man can row at 24 km/h, and run at 7 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 25 km, and the water is 24 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.

I've been stuck on this question for quite sometime now.. would anybody kindly help me out?

Thank you!
You are dealing with a right triangle:

The hypotenuse has the length 25, one leg has the length 24 and therefore the second leg must be 7 (Pythagorean theorem)

Since you know all three sides of the triangle you can choose the trigonometric function you like. I'll take Cosine function:

$\cos(\theta) = \dfrac{24}{25}~\implies~\theta\approx 16.26^\circ$

$\cos(\theta) = \dfrac{24}{25}~\implies~25=\dfrac{24}{\cos(\theta) }$

The definition of speed is:

$speed=\dfrac{distance}{time}~\implies~time=\dfrac{ distance}{speed}$

Plug in the values you already know:

$t = \dfrac{25}{24}$ that means you'll get: $t = \dfrac{\dfrac{24}{\cos(\theta)}}{24} = \dfrac1{\cos(\theta)} = \sec(\theta)$

I'm a little bit confused about the running speed of the man. You don't need this fact. If - and only if - you are asked to find the minimum time it takes to go from A to B by rowing and running, then, of course, my calculations are wrong.