Question

1. a helicopter takes off, and travels forward at an angle of 65.1° above horizontal. after following this path for 488 meters, the pilot changes the angle of flight to 12.5° above horizontal, and follows this path for 272 meters. after these two legs, what is the helicopters horizontal distance from the point of take off? 471 m 709 m 502 m 264 m

Explanation:

Step1: Calculate horizontal distance for first leg

The horizontal component of a vector \(d\) at an angle \(\theta\) is \(d\cos\theta\). For the first leg, \(d = 488\) meters and \(\theta=65.1^{\circ}\). So, \(x_1 = 488\cos(65.1^{\circ})\).
Using a calculator, \(\cos(65.1^{\circ})\approx0.420\), then \(x_1=488\times0.420 = 204.96\) meters.

Step2: Calculate horizontal distance for second leg

For the second leg, \(d = 272\) meters and \(\theta = 12.5^{\circ}\). The horizontal component \(x_2=272\cos(12.5^{\circ})\).
Using a calculator, \(\cos(12.5^{\circ})\approx0.976\), then \(x_2=272\times0.976=265.472\) meters.

Step3: Sum the two horizontal distances

The total horizontal distance \(x=x_1 + x_2\).
\(x=204.96+265.472 = 470.432\approx471\) meters.

Answer:

471 m