Question

altitude of 10,000 feet?

1. navigation an airplane flying at 600 miles per

hour has a bearing of 52°. after flying for 1.5 hours,
how far north and how far east will the plane have
traveled from its point of departure?

Explanation:

Step1: Calculate the distance traveled

The speed of the airplane is \(v = 600\) miles per hour and the time \(t=1.5\) hours. Using the formula \(d = vt\), we have \(d=600\times1.5 = 900\) miles.

Step2: Find the distance in the north - direction

If the bearing is \(\theta = 52^{\circ}\), the distance in the north - direction \(y\) (using the cosine function in a right - triangle where the hypotenuse is the distance traveled \(d\)) is \(y = d\cos\theta\). Substituting \(d = 900\) and \(\theta=52^{\circ}\), we get \(y = 900\cos(52^{\circ})\). Since \(\cos(52^{\circ})\approx0.6157\), \(y\approx900\times0.6157 = 554.13\) miles.

Step3: Find the distance in the east - direction

The distance in the east - direction \(x\) (using the sine function in a right - triangle where the hypotenuse is the distance traveled \(d\)) is \(x = d\sin\theta\). Substituting \(d = 900\) and \(\theta = 52^{\circ}\), we get \(x=900\sin(52^{\circ})\). Since \(\sin(52^{\circ})\approx0.7880\), \(x\approx900\times0.7880=709.2\) miles.

Answer:

The plane has traveled approximately \(554.13\) miles north and \(709.2\) miles east.