Question

(a) how far is the pilot from his destination at the time he notices the error? (round your answer to one decimal place.) enter a number. x mi (b) what bearing should he head his plane in order to arrive at airport b? (round your answer to the nearest degree.) n x° e

Explanation:

Step1: Calculate the distance flown east

The pilot flies east at a speed of \(225\) mi/h for \(40\) minutes (\(\frac{40}{60}=\frac{2}{3}\) hours).
Using the formula \(d = vt\) (distance = speed×time), the distance \(d = 225\times\frac{2}{3}=150\) mi.

Step2: Use the Law of Cosines for part (a)

Let \(a = 150\), \(b = 300\), and \(C = 50^{\circ}\).
The Law of Cosines is \(c^{2}=a^{2}+b^{2}-2ab\cos C\).
Substitute the values: \(c^{2}=150^{2}+300^{2}-2\times150\times300\times\cos50^{\circ}\).
First, calculate \(150^{2}=22500\), \(300^{2}=90000\), and \(2\times150\times300\times\cos50^{\circ}\approx2\times150\times300\times0.6428 = 57852\).
Then \(c^{2}=22500 + 90000-57852=54648\).
Take the square root: \(c=\sqrt{54648}\approx233.8\) mi.

Step3: Use the Law of Sines for part (b)

Let the angle opposite to the side of length \(150\) be \(\theta\).
By the Law of Sines \(\frac{\sin\theta}{150}=\frac{\sin50^{\circ}}{233.8}\).
\(\sin\theta=\frac{150\times\sin50^{\circ}}{233.8}\).
\(\sin50^{\circ}\approx0.7660\), so \(\sin\theta=\frac{150\times0.7660}{233.8}\approx0.492\).
\(\theta\approx29.5^{\circ}\).
The bearing is \(90^{\circ}-29.5^{\circ}=60.5^{\circ}\) (approx \(61^{\circ}\) when rounded to the nearest degree).

Answer:

(a) \(233.8\) mi; (b) \(61^{\circ}\)