Solve the problem.
-Radio direction finders are set up at points $A$ and $B , 8.68 \mathrm { mi }$ apart on an east-west line. From $A$ it is found that the bearing of a signal from a transmitter is $\mathrm { N } 54.3 ^ { \circ } \mathrm { E }$, while from $\mathrm { B }$ it is $\mathrm { N } 35.7 ^ { \circ } \mathrm { W }$. Find the distance of the transmitter from B, to the nearest hundredth of a mile.
A) $5.07 \mathrm { mi }$
B) $4.57 \mathrm { mi }$
C) $7.05 \mathrm { mi }$
D) $7.55 \mathrm { mi }$

Question

Quizplus · Accepted Answer

To solve this problem, we can use the law of sines in a triangle formed by points A, B, and the transmitter's location. The angle at A is $54.3^\circ$, and the angle at B is $35.7^\circ$. The angle opposite the side of length 8.68 mi (the distance between A and B) can be found by subtracting the sum of the two given angles from 180°, since the sum of angles in a triangle is 180°. Thus, the angle opposite the side AB is $180^\circ - 54.3^\circ - 35.7^\circ = 90^\circ$, indicating that the triangle is a right triangle with the right angle at the transmitter's location. The distance from B to the transmitter can be found using trigonometry (specifically, the cosine of 35.7°) as follows: $\cos(35.7^\circ) = \frac{8.68}{\text{distance from B to transmitter}}$, solving for the distance gives approximately $5.07 \mathrm { mi }$.

Solve the Problem AAA And B,8.68miB , 8.68 \mathrm { mi }B,8.68mi

Solve the Problem $A$ And $B , 8.68 \mathrm { mi }$