The pilot of an aircraft at a point A observes a mountain peak P on a bearing of `40^o ` at a distance 50 km. The aircraft flies on a bearing of `80^o ` at a constant height equal to that of the...

The pilot of an aircraft at a point A observes a mountain peak P on a bearing of `40^o ` at a distance 50 km. The aircraft flies on a bearing of `80^o ` at a constant height equal to that of the peak for a distance of 100 km to a point Q. Calculate:

(a) The distance PQ

(b) the angle PQA

(C) find the bearing of P from Q

Asked on by nyanav

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mathsworkmusic | (Level 2) Educator

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We have three points on a flat plane (above the ground, at the height of the mountain upon whose peak point P is situated): A, P, Q. Point A is the original position of the aircraft, point P is at the peak of the mountain and point Q is the point the aircraft is flying to.

We are told that the point P is at a bearing of `40^o` (from North) from A, 50km away. Also, we are told that Q (where the plane is headed) is at a bearing of `80^o` from A, 100km away.``

a) Find the distance PQ.

To do this, draw a triangle with vertices at A, P and Q. The angle opposite the side PQ, `Delta PAQ`  is the difference between the bearing of Q from A and the bearing of P from A, that is `Delta PAQ = 40^o` . We know also that the sides AP and AQ have lengths 50km and 100km respectively.

To find the length PQ consider the 3 knowns that we have about the triangle - 2 lengths and the angle between them. This should be recognisable as the scenario where you use the cosine rule:

`a^2 = b^2 + c^2 - 2bc cos A`

Here, let ``  `a = PQ`  (the length to be found), `` `b = AP` , `c = AQ`  and the angle opposite a, `A = Delta PAQ` . Then we have that

`PQ^2 = 50^2 + 100^2 - (50)(100)cos(40)`

Putting this into a calculator we get that

`PQ^2 = 2500 + 10000 - 5000 x 0.7660 = 8669.778 = 8670`

so that

`` km to 2dp

b) Find the angle `Delta PQA` .

We know from a) the angle and opposite side pair PQ and `Delta PAQ`  

and we require another angle whose opposite side we know the length of. This scenario should be recognised as requiring use of the sine rule:

`(sin A)/a = (sin B)/b = (sin C)/c`

Let `a = PQ = 93.1164`  and its opposite angle A `= Delta PAQ = 40^o` , and let `b = AP = 50` km as before and its opposite angle B be the one we wish to calculate, so that `B = Delta PQA` . Using the sine rule we have that

`(sin 40)/93.1164 = sin(B)/50`    which when rearranged gives that

`B = sin^(-1){(50/93.1164) sin 40} = sin^(-1)(0.3451704) = 20.1922 = 20.19^o ` to 2dp

That is `Delta PQA = 20.19^o`  to 2dp

c) Find the bearing of P from Q.

To find this bearing, draw another length to a point R from A so that a right-angled triangle AQR is formed. Using the rule of Z-angles, angle ```` `Delta AQR =` the bearing of Q from A, ie  `Delta AQR = 80^o` . From this it can be seen that `Delta AQR` , `Delta PQA`  and `360 - qp` , where `qp` is the bearing of P from Q, fall on a straight line so that they add to `180^o` . Therefore we can write that

`Delta AQR + Delta PQA + 360 - qp = 180`

We wish to find the bearing `qp`, so we can rearrange this equation to find that

`qp = Delta AQR + Delta PQA + 360 - 180 = 80 + 20.1922 + 180 = 280.1922`

So that the bearing of P from Q (from North) is `280.19^o` to 2dp

Sources:

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