A helicopter leaves a point P and flies on a bearing of 162 deg for 9 km to a point Q, and then due west to a point R which is on a bearing of 220 deg from P. i) find PR ii) find RQ The flight takes 4 minutes iii) find the speed of the helicopter in kilometers per hour. please help i really cant do this and i don't really understand the bearings at at i don't know how they work in comparison to trying to draw a diagram with angles...? if you could answer and show how you did it it would be a huge help :) thanks. Created by Stacey. education-reference-en - homework-help-en

maths a level help (trig)?

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maths a level help (trig)?

A helicopter leaves a point P and flies on a bearing of 162 deg for 9 km to a point Q, and then due west to a point R which is on a bearing of 220 deg from P.

i) find PR

ii) find RQ

The flight takes 4 minutes

iii) find the speed of the helicopter in kilometers per hour.

please help i really cant do this and i don't really understand the bearings at at i don't know how they work in comparison to trying to draw a diagram with angles...? if you could answer and show how you did it it would be a huge help :) thanks

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Answers & Comments

Maverick

Verified answer

If you're having trouble understanding it, draw it out.

Bearings denote the angle of travel clockwise around from North.

So start off drawing a point and labelling it P.

Then draw a small vertical straight line out of P - that's your North.

Measure 162 degrees clockwise from that line... mark that off and draw another line starting at P, at this angle, for the 9km distance to point Q.

Mark off point Q.

The helicopter then flies due West, which is to the left.

From point Q, draw a long faint horizontal line to the left. Somewhere along that line is point R.

Now go back to point P, and do the same as you did before. Measure 220 degrees from the little vertical line... mark off the 220 degree angle and draw another faint line starting at point P such that it crosses the original faint line. Where the two faint lines meet, that's point R.

Whatever you're doing, always think of a little line pointing North, and the bearing being the angle clockwise from that little line.

The angle at the top of this triangle must be 220 - 162 = 58 degrees

If you've drawn your rectangle, now drop a line from point P directly downward.

What happens? Now you see the line PQ splits off at 18 degrees from the horizontal, because your new line gives you a 180 degree angle compared to the 162 degree angle, and 180 - 162 = 18

You also see a right angle where your new line meets the line QR. That must be a 90 degree angle. What's left? The third angle in that sub-triangle, which is the angle at Q.

180 - 18 - 90 = 72

The angle PQR must be 72 degrees, and therefore the last angle must be 50.

First, we know a couple of angles, and one side, so use the sine rule:

a / sinA = b / sinB

9 / sin50 = PR / sin72

9*sin72 / sin50 = PR

PR = 11.173...

and again:

9 / sin50 = RQ / sin58

9*sin58 / sin50 = RQ

RQ = 9.963...

How far has the helicopter travelled? PQ and QR.