Question

HELPP!!!!

Answer 1

9514 1404 393

Answer:

  21 ≤ x < 48.3343

  19.4030 < t ≤ 62 1/3

Step-by-step explanation:

This sort of geometry has shown up in a number of different problems. Perhaps the most difficult was one in which the angles were given a specific linear relationship. Here, one of the angles is fixed, making the problem somewhat simpler.

The most straightforward approach is to simply say the unknown angle must lie between 0° and 180°. It turns out that is not the possible range of values. Rather there is a limit on the unknown angle. It will have an extreme value when both triangles are isosceles. The lengths of the equal sides in that case can be determined using the triangle with both angle and base length specified.

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a) For the triangle to be isosceles (MA = MH = MY), the side lengths must satisfy ...

  AM·sin(62°/2) = 7.1/2

  AM = 3.55/sin(31°) ≈ 6.89269

At that point, angle HMY will satisfy ...

  AM = (6.33/2)/sin((2x -42)°/2)

  3.55/sin(31°) = 3.165/sin(x-21°)

  (x -21)° = arcsin(3.165/3.55·sin(31°)) ≈ 27.3343°

  x = 48.3343

Because angle HMY can be smaller than this value, its minimum is 0:

  2x -42 = 0

  x -21 = 0

  x = 21

That is, x can range between 21 and 48.3343.

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b) Using similar logic on the other triangles, we find the unknown angle has an extreme value when the triangles are isosceles: CA = CB = CD = CD. As above, this will occur when ...

  (3t -7)° = 2·arcsin(17/16·sin(24°)) ≈ 51.2091°

  t = 19.4030

In this geometry, because AB > DE, the angle can only get larger than 51.2091°. Its upper limit is 180°, meaning that ...

  3t -7 ≤ 180

  3t ≤ 187

  t ≤ 62 1/3

In this geometry, the value of t can range between 19.4030 and 62 1/3.

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Additional comment

As we said at the beginning, your computer may be expecting you to come up with an answer that makes the unknown angle lie between 0 and 180°. For that range of angles, you have ...

  0 ≤ 2x-42 ≤ 180  ⇒  0 ≤ x -21 ≤ 90  ⇒  21 ≤ x ≤ 111

  0 ≤ 3t -7 ≤ 180  ⇒  0 ≤ t -7/3 ≤ 60  ⇒  7/3 ≤ t ≤ 187/3