Questions: A helicopter flies from the airport on a course with a bearing of 19°. After flying for 95 miles, the helicopter flies due east for some time. The helicopter flies back to the airport with a bearing of 227°. How far did the helicopter fly on the final leg of its journey? The distance the helicopter flew was approximately miles. (Do not round until the final answer. Then round to the nearest tenth.)

Solution Steps

Given the bearings:

• \( \text{bearing\_initial} = 19^\circ \)
• \( \text{bearing\_return} = 227^\circ \)

We can find the angles in the triangle:

• \( \angle A = 180^\circ - \text{bearing\_initial} = 180^\circ - 19^\circ = 161^\circ \)
• \( \angle B = \text{bearing\_return} - 180^\circ = 227^\circ - 180^\circ = 47^\circ \)
• \( \angle C = 180^\circ - (\angle A + \angle B) = 180^\circ - (161^\circ + 47^\circ) = -28^\circ \)

Convert the angles to radians for calculations:

• \( \angle A_{\text{rad}} = \frac{161 \pi}{180} \approx 2.80998 \)
• \( \angle B_{\text{rad}} = \frac{47 \pi}{180} \approx 0.82030 \)
• \( \angle C_{\text{rad}} = \frac{-28 \pi}{180} \approx -0.48869 \)

Using the Law of Sines: \[ \frac{\sin(\angle A)}{d_1} = \frac{\sin(\angle B)}{d_2} \] where \( d_1 = 95 \) miles (the initial distance) and \( d_2 \) is the distance flown due east.

Rearranging gives: \[ d_2 = d_1 \cdot \frac{\sin(\angle B)}{\sin(\angle A)} = 95 \cdot \frac{\sin(47^\circ)}{\sin(161^\circ)} \approx 213.4072 \text{ miles} \]

Using the Law of Cosines: \[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \] where:

• \( a = 95 \)
• \( b = 213.4072 \)
• \( C = -28^\circ \)

Calculating gives: \[ c^2 = 95^2 + (213.4072)^2 - 2 \cdot 95 \cdot 213.4072 \cdot \cos(-28^\circ) \] This results in: \[ c \approx 137.0 \text{ miles} \]

Final Answer