Vectors problem solving

Mathforum

A pilot wishes to fly from his home field to a destination 625km S20E. The cruising speed of the aircraft is 535 km/h. If there is a wind of 72km/h blowing from N80W, what heading should the pilot take in order to reach his destination and how long will the flight take (to the nearest minute)?

11rdc11

A pilot wishes to fly from his home field to a destination 625km S20E. The cruising speed of the aircraft is 535 km/h. If there is a wind of 72km/h blowing from N80W, what heading should the pilot take in order to reach his destination and how long will the flight take (to the nearest minute)?
Does S20E mean 20 degrees South of East.

Mathforum

S20E means south and 20 degrees east, or 70 degrees south of east

Unknown008

If you make a sketch, you see that the heading should be more towards the south and the angle between the straight line to the destination and the line showing the wind is 90-(10+20) = 60 degrees.

Now, use the sine rule...

$$\displaystyle \frac{sin(A)}{72} = \frac{sin(60)}{535}$$

You should get A = 6.69 degrees.

Therefore, the heading is 20 - 6.69 = 13.3 degrees, and in your notation, S13.3E.

Then, find the resultant speed. To do this, find the last angle of the triangle.
Angle = 180 - (60+6.69) = 113.7 degrees.

Then either by sine rule or cosine rule, find the last side of the triangle.
I'll use the cosine rule.

$$\displaystyle Speed = \sqrt{535^2 + 72^2 - 2(535)(72)cos(113.7)}$$

Speed = 567 km/hr.

Hence, time of travel = 625/567 = 1.10 hour or 1 hour 6 minutes

11rdc11
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