diff --git a/Grade 10/Math/MPM2DZ/Trig Quiz 1.md b/Grade 10/Math/MPM2DZ/Trig Quiz 1.md
index 6a4b93f..26275ca 100644
--- a/Grade 10/Math/MPM2DZ/Trig Quiz 1.md	
+++ b/Grade 10/Math/MPM2DZ/Trig Quiz 1.md	
@@ -1,67 +1,45 @@
 # Question 1
 
-```math
+$`\because \angle B^\prime = \angle B \quad (\text{Corresonding Line theorem})`$
 
-\because \angle B^\prime = \angle B \quad (\text{Corresonding Line theorem}) \\
+$`\because \angle C^\prime = \angle C \quad (\text{Corresponding Line theorem})`$ 
 
-\because \angle C^\prime = \angle C \quad (\text{Corresponding Line theorem}) \\
+$`\therefore \triangle AB^\prime C^\prime \sim \triangle ABC  \quad (\text{ AA } \sim) `$
 
-\therefore \triangle AB^\prime C^\prime \sim \triangle ABC  \quad (\text{ AA } \sim) \\ \\
+$`\therefore \dfrac{AB^\prime}{B^\prime C^\prime} = \dfrac{AB}{BC} `$
 
-\therefore \dfrac{AB^\prime}{B^\prime C^\prime} = \dfrac{AB}{BC} \\ 
+$`\therefore \dfrac{30}{14} = \dfrac{30+x}{22}`$
 
-\quad \\ 
+$`14(30+x) = 22(30) `$
 
+$`x = \dfrac{22(30)}{14} - 30 `$
 
-\therefore \dfrac{30}{14} = \dfrac{30+x}{22} \\ \\
+$`x = 17.1428571 \approx 17.14 `$
 
-\quad \\ 
+$`\dfrac{AC^\prime}{B^\prime C^\prime} = \dfrac{AC}{BC} `$
 
-14(30+x) = 22(30) \\
+$`\dfrac{y}{14} = \dfrac{y+15}{22} `$
 
-\quad \\
+$`22y = 14y + 14(15) `$
 
-x = \dfrac{22(30)}{14} - 30 \\
+$`8y = 14(15) `$
 
-\quad \\
-
-x = 17.1428571 \approx 17.14 
-```
-
-```math
-
-\dfrac{AC^\prime}{B^\prime C^\prime} = \dfrac{AC}{BC} \\
-
-\quad \\
-
-\dfrac{y}{14} = \dfrac{y+15}{22} \\
-
-\quad \\
-
-22y = 14y + 14(15) \\
-
-8y = 14(15) \\
-
-y = 26.25
-
-```
+$`y = 26.25`$
 
 # Question 2
 
-```math
+$`h = b \sin A`$
 
-h = b \sin A \\
+$`h = 11.3 \sin 32`$
 
-h = 11.3 \sin 32 \\
+$`h = 5.99`$
 
-h = 5.99 \\
+$`\because h \le 6.8 \le 11.3`$
 
-\because h \le 6.8 \le 11.3 \\
+$`\therefore 2 \triangle 's \text{ exist}`$
 
-\therefore 2 \triangle 's \text{ exist} \\
+$`\text{ Lets call point } T  \text{ is the height that is perpendicular on side } AB \text{ and connects to point } C. \text { and } B^\prime \text{ be the other possible point of } B.`$
 
-\text{ Lets call point } T  \text{ is the height that is perpendicular on side } AB \text{ and connects to point } C. \text { and } B^\prime \text{ be the other possible point of } B.
-```
 -------------------------
 $`\text{ Case } 1:`$
 
@@ -71,18 +49,18 @@ $`\angle CB^\prime T = 61.75^o`$
 
 $`\angle AB^\prime C = 180 - 61.75 = 118.25^o (\text{ Complentary Angle Theorem})`$
 
-$`\angle ACB^\prime = 180 - 61.75 - 32 = 86.25^o (\text{ASTT})`$
+$`\angle ACB^\prime = 180 - 118.25 - 32 = 29.75^o (\text{ASTT})`$
 
 $`\dfrac{AB}{\sin \angle ACB^\prime} = \dfrac{CB^\prime}{\sin A}`$
 
-$`\dfrac{AB}{\sin86.25} = \dfrac{6.8}{\sin 32}`$
+$`\dfrac{AB}{\sin29.75} = \dfrac{6.8}{\sin 32}`$
 
-$`AB = \dfrac{\sin 86.25 \times 6.8}{\sin32}`$
+$`AB = \dfrac{\sin 29.75 \times 6.8}{\sin32}`$
 
-$`AB = 12.8`$
+$`AB = 6.37`$
 
 -------------------------
-$`\text{ Case} 2: `$
+$`\text{ Case } 2: `$
 
 $`\angle ABC = 61.75`$
 
@@ -92,4 +70,26 @@ $`\dfrac{AB}{\sin C} = \dfrac{CB}{\sin A}`$
 
 $`\dfrac{AB}{\sin 86.25} = \dfrac{6.8}{\sin 32}`$
 
-$`AB = 
\ No newline at end of file
+$`AB = \dfrac{\sin 86.25 \times 6.8}{\sin 32}`$
+
+$`AB = 12.8`$
+
+# Question 3
+
+$`\text{let the square be } ABCD \text{ and the inner triangle } AEF `$
+
+$`\sin (\beta) = \dfrac{EF}{AE} = \dfrac{EF}{1} = EF`$
+
+$`\sin (\alpha) \sin(\beta) = \dfrac{EF}{AE} \times \dfrac{EC}{EF} = \dfrac{EC}{AE} = \dfrac{EC}{1} = EC`$
+
+$`\cos(\alpha) \sin(\beta) = \dfrac{CF}{EF} \times \dfrac{EF}{AE} = \dfrac{CF}{AE} = \dfrac{CF}{1} = CF`$
+
+$`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \sin(\alpha + \beta) = \dfrac{CD}{AE} = \dfrac{CD}{1} = CD`$
+
+$`\cos(\alpha) \cos(\beta) = \dfrac{AD}{AF} \times \dfrac{AF}{AE} = \dfrac{AD}{1} = AD`$
+
+$`\sin(\alpha) \cos(\beta) = \dfrac{FD}{AF} \times \dfrac{AF}{AE} = \dfrac{FD}{1} = FD`$
+
+$`\cos(\beta) = \dfrac{AF}{AE} = \dfrac{AF}{1} = AF`$
+
+$`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \cos(\alpha + \beta) = \dfrac{BE}{AE} = \dfrac{BE}{1} = BE`$