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9514 1404 393

  3 < x < 9.09593

Step-by-step explanation:

In the attached, we defined TU=TW=y and TV=z. Then we used the Law of Cosines to write equations involving x, y, and z:

  42² = y² +z² -2yz·cos((6x-18)°)

  44² = y² +z² -2yz·cos((4x+2)°)

Solving each of these quadratics for z, and setting those solutions equal, gives a relation between x and y. We impose the constraints that x > 3, y > 0, and z > 0 so that angles and segment lengths are non-negative.

The attached graphs show the relation has solutions in the range of ...

  3 < x < 9.09593 (approximately)

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Comment on the question

Perhaps there is a simple way to solve this, but I haven't seen it yet. There is a set of constraints on x imposed by ∠UVW ≤ 180° and by ∠UTV ≥ 0°, but it turns out the geometry is actually more restrictive than that. The hard part of the solution is maintaining the required relation between angles VTU and VTW.

Added comment

The problem is symmetrical in that y and z are interchangeable. As the graph in the attachment shows, they converge to a single value at the extreme value of x. That is, the maximum x can be found by solving the equation ...

  (1 -cos((6x-18)°))/42² = (1 -cos((4x+2)°)/44²

That solution is x ≈ 9.09592829977.

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step-by-step explanation:

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step-by-step explanation: