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## Question

A plane moves from point 0 and travels in a straight horizontal line a distance of X meters

before it flies a distance of 17X in the same direction making with the earth line an angle X degrees.

Then in the moment the pilot decides to maintain a constant latitude he turns anticlockwise with a certain angle and radius,

and flies horizontally in the new direction for a distance of 18X before he lands to the starting point 0 by an angle of 3X/2.

I think that's all the info needed. Find X.

## 13 answers to this question

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The description of events is a bit unclear, especially the last part "flies horizontally in the new direction for a distance of 18X before he lands to the starting point 0 by an angle of 3X/2." A diagram might help. Thanks.

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X = 45

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The description of events is a bit unclear, especially the last part "flies horizontally in the new direction for a distance of 18X before he lands to the starting point 0 by an angle of 3X/2." A diagram might help. Thanks.

this is a top view diagram.. hope it makes it easier to understand.

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well.. i didn't think it would be that hard...

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Clarification? The problem states the plane starts at 0 and goes straight for X meters. However, the provided picture shows 12X. Which is correct?

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It does seem that there are some inconsistencies here

as I read the problem, I'm not sure that they are material. Maybe there are just two triangles to compare and the rest is red herring. Both triangles start at ground level and reach the same height. One triangle with an angle(X) and hypotenuse 17X and the other with an angle(3X/2) and hypotenuse 18X. Then 17sin(X)=18sin(3X/2) and I get X=75.805o? Kind of hard to visualize a plane climbing at an angle for a prescribed distance, circling at a constant elevation, and then descending at a steeper angle over a greater distance than the initial climb and the result not be affected by the distance between the point of initial climb and the endpoint though so I suspect I'm way off .

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Clarification? The problem states the plane starts at 0 and goes straight for X meters. However, the provided picture shows 12X. Which is correct?

Sorry for this mismatch.. 12X is the correct distance

The diagram can be very helpful (top view)

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It does seem that there are some inconsistencies here

as I read the problem, I'm not sure that they are material. Maybe there are just two triangles to compare and the rest is red herring. Both triangles start at ground level and reach the same height. One triangle with an angle(X) and hypotenuse 17X and the other with an angle(3X/2) and hypotenuse 18X. Then 17sin(X)=18sin(3X/2) and I get X=75.805o? Kind of hard to visualize a plane climbing at an angle for a prescribed distance, circling at a constant elevation, and then descending at a steeper angle over a greater distance than the initial climb and the result not be affected by the distance between the point of initial climb and the endpoint though so I suspect I'm way off .

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x = arccos(11/12) = 23.55646430917

Any chance we can see your solution + work?

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this is a top view diagram.. hope it makes it easier to understand.

working in 2D will make it much easier

reference to the sketch, and knowing that two tangents fom one point to the same circle have the same length,

then, regardless to the radius of the circle, the two horizontal projections (which are the two tangents) are of the same length

which means 12x + 17x(COSx) = 18x + N COS(3x/2) [eq.1]

N is the length of the last portion (landing back to 0) having the angle withn the ground 3x/2

to find N, we consider the plane reaches to a constant height, and keeps horizontal till landing,

which means 17x SINx = N SIN(3x/2) => N = 17x SINx / SIN(3x/2)

in [eq.1], 6x + 17x (SINx COT(3x/2) - COSx) = 0

this has certailnly a solution x=0, but as we need a different solution,

we consider x > 0 and 17(SINx COT(3x/2) - COSx) + 6 = 0

to find x we... or maybe i'll stop here to see wat u have to say abt this.

FIND X.

17(SINx COT(3x/2) - COSx) + 6 = 0

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3ala2 - I agree with your work...

arccos(11/12) is a solution to your halfway equation

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3ala2 - I agree with your work...

arccos(11/12) is a solution to your halfway equation

Good job ljb, this is the correct answer.

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Edited by 3ala2

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