find the distance and degree

Question

You leave your friend behind on the shore and you travel 3 miles due east in your boat. Then you travel 2 miles northeast. Then you travel 1 mile due north. Your friend can see you at a distance of BLANK miles and at a bearing of BLANK degrees

use the law of sines and cosines

Answer 1

We need bearings and trigonometry for this one. Bearings are measured clockwise from North (0 deg), so East is 90 deg.

First we should draw a sketch so we know what the problem looks like.

Start at point A, with your friend. Draw a line due East (bearing 90 deg) to represent 3 miles to point B. Then draw a line North-East (bearing 45 deg) to represent 2 miles to point C. Then draw line CA, which we need to find (both its length and bearing from A).

We have a triangle, and we have 2 known sides, and we need to find the other side. So we need to find at least one internal angle before we can solve it.

I think the easiest is angle ABC because this is formed from East and North-East lines. This tells us that the angle is 90 + 45 = 135 deg.

We can immediately use the cosine rule to solve: c = √ (a² + b² - 2ab cos C)

So that CA = √ (3² + 2²) - 2(3)(2) cos (135 deg) )

=> CA = 4.64 miles (2dp)

Now we have the third side, we can use the sine rule:

sin A / a = sin B / b = sinc C / c

so that to find the angle CAB we can see that sin (135 deg) / 4.64 = sin CAB / 2

=> 2 sin (135 deg) / 4.64 = sin CAB

=> arcsin ( 2 sin (135 deg) / 4.64 ) = CAB = 17.745 deg or 18 deg (to nearest deg)

Hence, bearing is 90 - 18 = 72 deg

Our friend can see you at a distance of 4.64 miles and at a bearing of 72 degrees.

If you want to use formulae for right angle triangles only, you can, but it takes 2 steps...

First, add another point, Z, to your sketch. This is due South of C, and due East of A and B. This make a small right angle triangle BZC, and a large one AZC.

Starting with BZC, we must find the side BZ (as we already have AB = 3).

cos (45 deg) = BZ / hypotenuse

=> cos (45 deg) x hypotenuse = BZ

=> 2 cos (45 deg) = BZ = 1.414 (= √2)

sin (45 deg) = ZC / hypotenuse

=> sin (45 deg) x hypotenuse = ZC

=> 2 sin (45 deg) = ZC = 1.414 (= √2)

Now we can use big triangle AZC to find angle CAZ and length CA (the hypotenuse).

tan CAZ = opp / adj = √2 / (3 + √2)

CAZ = arctan ( √2 / (3 + √2) ) = 17.764 = 18 degrees (to nearest deg) and bearing = 90-18 = 72 deg

and using pythagoras

CA² = AZ² + CZ² = (3 + √2)² + (√2)²

CA = √ ( (3 + √2)² + (√2)² ) = 4.635 or 4.64 to 2 dp