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Solve the triangle. A = 19°, C = 102°, c = 6
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i got B = 59°, a ≈ 2, b ≈ 5.3
just checking my answer
59 is of course correct
\[\frac{6}{\sin(102)}=\frac{a}{\sin(19)}\] so \[a=\frac{6\sin(19)}{\sin(102)}\] is that about 2?
yeah looks like about 2
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yay
want me to check the last one? probably right as well
no thts okay thank u
I previously entered some figures here but I deleted those because I was thinking (incorrectly) this was two angles and an included side. 59° side 6 19° IS an ASA congruency problem and the remaining sides are: 5.2579 1.997
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