Solve the problem.
-Lookout station B is located $12 \mathrm { mi }$ due east of station A. The bearing of a fire from A is $\mathrm { S } 10 ^ { \circ } 10 ^ { \prime } \mathrm { W }$ and the bearing from B is $\mathrm { S } 38 ^ { \circ } 40 ^ { \prime } \mathrm { W }$. Determine the distance from the fire to $\mathrm { B }$ (to the nearest tenth of a mile).
A) $14.4 \mathrm { mi }$
B) $24.8 \mathrm { mi }$
C) $26.8 \mathrm { mi }$
D) $12.4 \mathrm { mi }$

Question

Quizplus · Accepted Answer

To solve this problem, we can use the law of sines in a triangle formed by the two stations and the fire. The angle at station A is $10^\circ 10'$, and the angle at station B is $38^\circ 40'$. The angle between station A and B, relative to the fire, can be found by subtracting the bearing of the fire from B from $180^\circ$ (since the bearings are given from the north and we need the internal angle of the triangle). This gives us $180^\circ - 38^\circ 40' = 141^\circ 20'$. The distance between A and B is 12 miles, which is the side opposite the angle we just found. Using the law of sines, we can find the distance from B to the fire. Let $a = 12$ miles (distance between A and B), $A = 141^\circ 20'$ (angle opposite side $a$), $B = 38^\circ 40'$ (angle at station B), and $b$ is what we're solving for (distance from B to the fire). $$\frac{a}{\sin A} = \frac{b}{\sin B}$$Solving for $b$, we get:$$b = \frac{a \sin B}{\sin A}$$Plugging in the values:$$b = \frac{12 \sin(38^\circ 40')}{\sin(141^\circ 20')}$$Calculating this gives a value close to $24.8$ miles, which is option B.

Solve the Problem 12mi12 \mathrm { mi }12mi Due East of Station A

Solve the Problem $12 \mathrm { mi }$ Due East of Station A