Based on the diagram shown, find θ to the nearest degree.

  θ = 38°

Step-by-step explanation:

The lower right triangle is congruent to the upper left triangle, so we have θ and 20° being the two acute angles in the triangle. The law of sines tells you ...

  sin(θ)/9 = sin(20°)/5

  sin(θ) = (9/5)sin(20°)

  θ = arcsin(9/5·sin(20°)) ≈ 38°

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Another solution to the triangle is θ = 180° -38° = 142°. The diagram clearly shows θ as an acute angle, so we take this second solution to be extraneous.

true? is that what you're asking or do you mean something else?