garydesir1
garydesir1 garydesir1
  • 01-07-2017
  • Mathematics
contestada

I should be able to solve this but some reasons I just can't see the steps and the right answer.

Can you help me ? Thanks!

I should be able to solve this but some reasons I just cant see the steps and the right answer Can you help me Thanks class=

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Banabanana
Banabanana Banabanana
  • 01-07-2017
[tex]sin(20^o)= \frac{opposite}{hypotenuse} = \frac{19}{x} \ \ \ \to \\ \\ x = \frac{19}{sin(20^o)}= \frac{19}{0.342} \approx 55.6[/tex]
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