Question

212 trigonometry lesson #8: further applications ... sine law & cosine law numerical 11. response
two aircraft x and y leave an airport at the same time. x flies on a course of 70° at 720 km/h, and y flies on a course of 350° at 600 km/h.
to the nearest kilometre, the distance between the aircraft after 5 minutes is
(record your answer in the numerical response box from left to right.)

Explanation:

Step1: Calculate the distances traveled by each aircraft

• Convert 5 minutes to hours: \(t=\frac{5}{60}=\frac{1}{12}\) hours.
• Distance formula \(d = vt\) (where \(v\) is velocity and \(t\) is time).
• For aircraft \(X\): \(d_X=720\times\frac{1}{12} = 60\) km.
• For aircraft \(Y\): \(d_Y=600\times\frac{1}{12}=50\) km.

Step2: Find the angle between their paths

• The angle between their courses \(\theta=70^{\circ}-350^{\circ}+360^{\circ}=80^{\circ}\) (using the concept of bearing angles).

Step3: Apply the cosine law

• The cosine law formula is \(c^{2}=a^{2}+b^{2}-2ab\cos C\), where \(a = d_X = 60\), \(b = d_Y=50\), and \(C=\theta = 80^{\circ}\).
• \(c^{2}=60^{2}+50^{2}-2\times60\times50\times\cos(80^{\circ})\).
• \(c^{2}=3600 + 2500-6000\times0.173648\).
• \(c^{2}=6100-1041.888\).
• \(c^{2}=5058.112\).
• \(c=\sqrt{5058.112}\approx71\) km.

Answer:

\(71\)