# Thread: Law of sines word problems

1. ## Law of sines word problems

a family is traveling due west on a road that passes a famous landmark. at a given time the bearing to the landmark is N 62 degrees W, and after the family travels 5 miles farther the bearing is N 38 degrees W. What is the closest the family will come to the landmark while on the road?

2. Originally Posted by jordangiscool
a family is traveling due west on a road that passes a famous landmark. at a given time the bearing to the landmark is N 62 degrees W, and after the family travels 5 miles farther the bearing is N 38 degrees W. What is the closest the family will come to the landmark while on the road?
let $d$ = closest (perpendicular) distance to the mark from the road

let $x$ = initial distance along the road to the closest point

$\frac{d}{x} = \tan(28)$

$\frac{d}{x-5} = \tan(52)$

two equations , two unknowns ... solve the system.

3. Originally Posted by jordangiscool
a family is traveling due west on a road that passes a famous landmark. at a given time the bearing to the landmark is N 62 degrees W, and after the family travels 5 miles farther the bearing is N 38 degrees W. What is the closest the family will come to the landmark while on the road?
Hi jordangiscool,

Here's another approach using the Law of Sines.

Refer to the diagram.

Starting at Point A traveling due West a Landmark L is sighted at a bearing of N 62 W which is 62 degrees to the West of North.

The complementary angle is 28 degrees.

After traveling 5 miles to Point B, the new bearing is N 38 W which is 38 degrees West of North.

The complementary angle here is 52 degrees.

We need to find the perpendicular distance from Point C to Landmark L.

Using a little geometry, the rest of the angles can be determined.

Now, refer to Triangle ABL. Using the Law of Sines we can determine the length of BL. Once we know BL, we can use simple right triangle trigonometry to find CL.

Finding BL:

$\frac{\sin 24}{5}=\frac{\sin 28}{BL}$

$BL=\frac{5\sin 28}{\sin 24}$

Finding CL:

$\sin 52=\frac{CL}{\frac{5\sin 28}{\sin 24}}$

$CL=\sin 52 \cdot \frac{5\sin 28}{\sin 24}\approx 4.5$ miles.