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education/math/MATH1060 (trig)/Law of Sines.md

@@ -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: $a = h$