# A driver leaves Toronto and heads towards Buffalo, via Niagra Falls. The driver checks the gpS device and records the following direction: Distance from Toronto to Niagra Falls: 1540.762km, with a...

A driver leaves Toronto and heads towards Buffalo, via Niagra Falls. The driver checks the gpS device and records the following direction:

Distance from Toronto to Niagra Falls: 1540.762km, with a direction S 35 °21'30 W

Distance from Niagra Falls to Buffalo: 330.762km, with a direction : S 3 °11' 20 W

If an alternative route were taken from Toronto to Buffalo avoiding Niagra Falls, would the distance be londer or shorter?

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**Answer: It is shorter.**

Not taking into account windy/mountainous roads etc, this is a straight vector addition/subtraction problem.

However, I am not familiar with the way the angle is represented so I will assume that the first angle refers to 35°21'30" to the west of due south.

The key to these is to draw a picture. Three points represent Toronto, Niagra Falls, and Buffalo. With Toronto being your origin, Niagra Falls is 1540.762 km 35°21'30" West of South of Toronto. This will be represented by vector `vec(TN)`

To convert 35°21'30 to an angle in decimal:

d°m's" => decimal conversion: `d+(m*60+s)/3600`

`35+(21*60+30)/3600=35.3583...^o`

And the other angle is 3°11'20" => `3.1889...^o`

So Buffalo is 3.1889 west of south of Niagra Falls. (Vector `vec(NB)` )

This is a vector addition problem, since we are looking to compare Toronto to Buffalo directly.

Evaluate the components of both vectors, add those components, and determine the resultant length. If that length is less than the two original vectors added, then travelling directly to Buffalo is shorter.

Vectors can be broken down into vector components by applying a bit of trigonometry:

Vector `vec(TN)`has a Southerly component of `TN*cos(35.3583)=1540.762text(km)*cos(35.3583)~~1256.57text(km)` and a Westerly component of `TN*sin(35.3583)=1540.762text(km)*sin(35.3583)~~891.62text(km)`

Vector `vec(NB)` has a Southerly component of `NB*cos(3.1889)=330.762text(km)*cos(3.1889)~~330.25text(km)` and a Westerly component of `NB*sin(3.1889)=330.762text(km)*sin(3.1889)~~18.40text(km)`

Adding the south components together, we get

`1256.57text(km)+330.25text(km)=1586.82text(km)`

And adding the west components together:

`891.62text(km)+18.40text(km)=910.02text(km)`

The length of the resultant is found using the pythagorean theorem:

`TB=sqrt((1586.82text(km))^2+(910.02text(km))^2)`

`TB=1829.24text(km)`

Compared to the two vector lengths added (distance): `1540.762text(km)+330.762text(km)=1871.52text(km)`

**ANSWER: Which means that bypassing Niagra Falls would be shorter by about 42.3 km**

Side Note: in reality Toronto to Niagra Falls 'as the crow flies' is about 62 km South East. Looking up GoogleMaps, the trip would only be different by about 4 km - you have to go around Lake Ontario. :)