Why isn't this triangle an ambiguous case? (where there can be 2 possible triangles from the same set of lengths and an angle)

Question

I've figured out the height of the triangle $h = 3.4202$ $h=3.4202$ and since that ambiguous cases happen when $h < a < b$ $h< a< b$ where $a$ $a$ is the length opposite of a given length and $b$ $b$ is some other length, shouldn't this triangle have 2 possible answers?

I'm terribly new to the whole subject of ambiguous cases so any help would be greatly appreciated!
Thanks in advance!

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Answer 1 · 2018-04-09T09:45:56

See below.

enter image source here

This is your triangle. As you can see it is an ambiguous case.

So to find the angle $θ$ $theta$ :

$\frac{sin (20^{\circ})}{8} = \frac{sin (θ)}{10}$ $sin(20^@)/8=sin(theta)/10$

$sin (θ) = \frac{10 sin (20^{\circ})}{8}$ $sin(theta)=(10sin(20^@))/8$

$θ = arcsin (\frac{10 sin (20^{\circ})}{8}) = {25.31}^{\circ}$ $theta=arcsin((10sin(20^@))/8)=color(blue)(25.31^@)$

Angles on a straight line add to $180^{\circ}$ $180^@$ , so other possible angle is:

$180^{\circ} - {25.31}^{\circ} = {154.69}^{\circ}$ $180^@-25.31^@=color(blue)(154.69^@)$

You can see from the diagram that, as you noted:

$h < a < b$ $h< a < b$

Here is a link that may help you. This can take a while to grasp, but you seem to be on the right track.