vault backup: 2024-11-11 21:48:16

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zleyyij 2024-11-11 21:48:16 -07:00
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@ -16,5 +16,18 @@ Under convention:
These triangles can be solved by adding a line that goes from one vertex to intersect perpendicular to the opposite side, forming two right triangles ($h$).
## Solving for the law of sines
We know that $\sin\alpha = \dfrac{h}{b}$
We know that $\sin\alpha = \dfrac{h}{b}$ and $\sin\beta = \dfrac{h}{a}$. We can sole both equations for $h$ to get:
- $h = b\sin\alpha$
- $h = a\sin\beta$
Setting both equations equal to each other gives us:
$b\sin\alpha = a\sin\beta$
Multiply both sides by $\dfrac{1}{ab}$ gives yields $\dfrac{\sin\alpha}{a} = \dfrac{\sin\beta}{b}$
# SSA triangles
Side side angle triangles may be solved to have one possible solution, two possible solutions, or no possible solutions.
- No triangle: $a < h$
- One triangle: $a \ge b$
- Two triangles: $h < a < b$
- One right triangle: $a = h$