diff --git a/Grade 10/Math/MPM2DZ/Trig Quiz 1.md b/Grade 10/Math/MPM2DZ/Trig Quiz 1.md
index 5b708b4..5c4809c 100644
--- a/Grade 10/Math/MPM2DZ/Trig Quiz 1.md	
+++ b/Grade 10/Math/MPM2DZ/Trig Quiz 1.md	
@@ -4,7 +4,7 @@ $`\because \angle B^\prime = \angle B \quad (\text{PLT-F})`$
 
 $`\because \angle C^\prime = \angle C \quad (\text{PLT-F})`$ 
 
-$`\therefore \triangle AB^\prime C^\prime \sim \triangle ABC  \quad (\text{ AA } \sim) `$
+$`\therefore \triangle AB^\prime C^\prime \sim \triangle ABC  \quad (\text{AA } \sim) `$
 
 $`\therefore \dfrac{AB^\prime}{B^\prime C^\prime} = \dfrac{AB}{BC} `$
 
@@ -16,9 +16,9 @@ $`x = \dfrac{22(30)}{14} - 30 `$
 
 $`x = 17.1428571 \approx 17.14 `$
 
-$`\dfrac{AC^\prime}{B^\prime C^\prime} = \dfrac{AC}{BC} `$
+$`\therefore \dfrac{AC^\prime}{B^\prime C^\prime} = \dfrac{AC}{BC} `$
 
-$`\dfrac{y}{14} = \dfrac{y+15}{22} `$
+$`\therefore \dfrac{y}{14} = \dfrac{y+15}{22} `$
 
 $`22y = 14y + 14(15) `$
 
@@ -49,7 +49,7 @@ $`\angle CB^\prime T = \sin^{-1} \Bigl(\dfrac{5.99}{6.8} \Bigr)`$
 
 $`\angle CB^\prime T = 61.75^o`$
 
-$`\angle AB^\prime C = 180 - 61.75 = 118.25^o (\text{ Complentary Angle Theorem})`$
+$`\angle AB^\prime C = 180 - 61.75 = 118.25^o (\text{CAT})`$
 
 $`\angle ACB^\prime = 180 - 118.25 - 32 = 29.75^o (\text{ASTT})`$
 
@@ -64,9 +64,9 @@ $`AB = 6.37`$
 -------------------------
 $`\text{ Case } 2: `$
 
-$`\angle ABC = 61.75`$
+$`\angle ABC = \angle CB^\prime T = 61.75^o (\text{ITT})`$
 
-$`\angle ACB = 180 - 32 - 61.75 = 86.25^o (\text{ ASTT})`$
+$`\angle ACB = 180 - 32 - 61.75 = 86.25^o (\text{ASTT})`$
 
 $`\dfrac{AB}{\sin C} = \dfrac{CB}{\sin A}`$
 
@@ -82,17 +82,17 @@ $`\therefore AB \text{ could either be } 6.37cm \text{ or } 12.8cm`$
 
 # Question 3
 
-$`\text{let the square be } \square ABCD \text{ and the inner triangle } \triangle AEF `$
+$`\text{let the square be } \square ABCD \text{ and the inner triangle be } \triangle AEF `$
 
 <img src="https://files.catbox.moe/ex7mea.png" width="1000">
 
 $`\sin (\beta) = \dfrac{EF}{AE} = \dfrac{EF}{1} = EF`$
 
-$`\sin (\alpha) \sin(\beta) = \dfrac{EF}{AE} \times \dfrac{EC}{EF} = \dfrac{EC}{AE} = \dfrac{EC}{1} = EC`$
+$`\sin (\alpha) \sin(\beta) = \dfrac{EC}{EF} \times \dfrac{EF}{AE} = \dfrac{EC}{AE} = \dfrac{EC}{1} = EC`$
 
 $`\cos(\alpha) \sin(\beta) = \dfrac{CF}{EF} \times \dfrac{EF}{AE} = \dfrac{CF}{AE} = \dfrac{CF}{1} = CF`$
 
-$`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \sin(\alpha + \beta) = \dfrac{CD}{AE} = \dfrac{CD}{1} = CD`$
+$`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \quad \sin(\alpha + \beta) = \dfrac{CD}{AE} = \dfrac{CD}{1} = CD`$
 
 $`\cos(\alpha) \cos(\beta) = \dfrac{AD}{AF} \times \dfrac{AF}{AE} = \dfrac{AD}{1} = AD`$
 
@@ -100,4 +100,16 @@ $`\sin(\alpha) \cos(\beta) = \dfrac{FD}{AF} \times \dfrac{AF}{AE} = \dfrac{FD}{1
 
 $`\cos(\beta) = \dfrac{AF}{AE} = \dfrac{AF}{1} = AF`$
 
-$`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \cos(\alpha + \beta) = \dfrac{BE}{AE} = \dfrac{BE}{1} = BE`$
+$`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \quad \cos(\alpha + \beta) = \dfrac{BE}{AE} = \dfrac{BE}{1} = BE`$
+
+## Footer Notes
+
+$`\text{ASTT} = \text{ Angle Sum Of Triangle Theorem}`$
+
+$`\text{ITT} = \text{ Isoceles Triangle Theorem}`$
+
+$`\text{CAT} = \text{ Corresponding Angle Theorem}`$
+
+$`\text{PLT-F} = \text{ Parallel Line Theorem; F-pattern}`$
+
+$`\text{AA} \sim \space = \text{ Angle Angle Similarity Theorem}`$
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