Question

the bearing from a to c is s 64° e. the bearing from a to b is n 82° e. the bearing from b to c is s 26° w. a plane flying at 250 mph takes 2.2 hr to go from a to b. find the distance from a to c. (round to the nearest integer as needed.) the distance from a to c is mi.

Explanation:

Step1: Calculate distance from A to B

We know speed \(v = 250\) mph and time \(t=2.2\) hr. Using the formula \(d = vt\), we have \(d_{AB}=250\times2.2 = 550\) mi.

Step2: Analyze the bearing - related angles

The bearing from A to B is N \(82^{\circ}\) E, and from A to C is S \(64^{\circ}\) E, and from B to C is S \(26^{\circ}\) W. We can find the angles in the triangle ABC. The angle \(\angle BAC=82^{\circ}+ 64^{\circ}=146^{\circ}\), and the angle \(\angle ABC = 82^{\circ}-26^{\circ}=56^{\circ}\), then the angle \(\angle ACB=180^{\circ}-146^{\circ}-56^{\circ}= - 22^{\circ}\), which is wrong. Let's correct the angle - finding. The angle \(\angle BAC = 180^{\circ}-(82^{\circ}+64^{\circ})=34^{\circ}\), the angle \(\angle ABC=82^{\circ}+26^{\circ}=108^{\circ}\), so \(\angle ACB=180^{\circ}-34^{\circ}-108^{\circ}=38^{\circ}\)

Step3: Use the Law of Sines

The Law of Sines states that \(\frac{d_{AB}}{\sin\angle ACB}=\frac{d_{AC}}{\sin\angle ABC}\). We know \(d_{AB} = 550\) mi, \(\angle ACB = 38^{\circ}\), \(\angle ABC=108^{\circ}\). So \(d_{AC}=\frac{d_{AB}\sin\angle ABC}{\sin\angle ACB}=\frac{550\times\sin108^{\circ}}{\sin38^{\circ}}\)

Step4: Calculate the value

\(\sin108^{\circ}\approx0.9511\), \(\sin38^{\circ}\approx0.6157\). Then \(d_{AC}=\frac{550\times0.9511}{0.6157}=\frac{523.105}{0.6157}\approx850\) mi

Answer:

850