Anybody good with using Law of Cosines? here's a problem for ya!?

Question

Navigation: A plane flies 810 miles from Niagara to Cuyahoga with a bearing of 75 degrees. Then it flies 648 miles to Chuyahoga to Rosemount with a bearing of 32 degrees. Draw a figure that visually represents the problem, and find the straight-line distance and bearing from Niagara to Rosemount.

Answer 1

Draw triangle NCR with NC=810, CR=648. Now, the angle between vectors C-N and R-C is 75°-32°, or 43°. However, note that since R-C is located at the _tip_ of C-N, rather than the tail, that this angle is in fact the exterior angle at C. Therefore, the interior angle (which we need for the law of cosines) is the supplement of 43°, which is 137°. Now, according to the law of cosines:

NR² = NC² + CR² - 2*NC*CR*cos θ
NR² = 810² + 648² - 2*810*648*cos 137°
NR² ≈ 1843750
NR ≈ 1357.8 miles

Let us check our work by using the method of components:

(C-N)_x = 810 cos 75° ≈ 209.6434
(R-C)_x = 648 cos 32° ≈ 549.5352
(R-N)_x = (C-N)_x + (R-C)_x ≈ 759.1786

(C-N)_y = 810 sin 75° ≈ 782.3999
(R-C)_y = 648 sin 32° ≈ 343.3877
(R-N)_y = (C-N)_y + (R-C)_y ≈ 1125.7876

|R-N| = √(((R-N)_x)² + ((R-N)_y)²)
|R-N| ≈ √1843750 ≈ 1357.8 miles

So we have indeed obtained the correct answer.

Answer 2

431.05 miles