From 4dcb55ab6cf5a314ead3364fb78d2424331cfcd9 Mon Sep 17 00:00:00 2001
From: zleyyij <75810274+zleyyij@users.noreply.github.com>
Date: Mon, 11 Nov 2024 21:48:16 -0700
Subject: [PATCH] vault backup: 2024-11-11 21:48:16

---
 education/math/MATH1060 (trig)/Law of Sines.md | 15 ++++++++++++++-
 1 file changed, 14 insertions(+), 1 deletion(-)

diff --git a/education/math/MATH1060 (trig)/Law of Sines.md b/education/math/MATH1060 (trig)/Law of Sines.md
index bf62b11..6a1c6ef 100644
--- a/education/math/MATH1060 (trig)/Law of Sines.md	
+++ b/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: $a = h$