Solve the problem.
-Radio direction finders are set up at points $\mathrm { A }$ and $\mathrm { B } , 7.24 \mathrm { mi }$ apart on an east-west line. From $A$ it is found that the bearing of a signal from a transmitter is $\mathrm { N } \mathrm {} 50.2 ^ { \circ } \mathrm { E }$, while from B it is $\mathrm { N } 39.8 ^ { \circ } \mathrm { W }$. Find the distance of the transmitter from $B$, to the nearest hundredth of a mile.
A) $4.13 \mathrm { mi }$
B) $6.06 \mathrm { mi }$
C) $4.63 \mathrm { mi }$
D) $5.56 \mathrm { mi }$

Question

Quizplus · Accepted Answer

The problem can be solved using trigonometry, specifically the law of sines. First, we need to find the angles of the triangle formed by points A, B, and the transmitter. The bearing from A is $50.2^\circ$ east of north, and from B, it is $39.8^\circ$ west of north. These bearings help us to form a triangle where we can calculate the internal angles and then use the law of sines to find the distance from B to the transmitter. The correct calculation leads to the distance being approximately $4.63$ miles, making option C the correct answer.

Solve the Problem A\mathrm { A }A And B,7.24mi\mathrm { B } , 7.24 \mathrm { mi }B,7.24mi

Solve the Problem $\mathrm { A }$ And $\mathrm { B } , 7.24 \mathrm { mi }$