Update Trig Quiz 1.md

Grade 10/Math/MPM2DZ/Trig Quiz 1.md

@@ -45,3 +45,51 @@ x = 17.1428571 \approx 17.14
 y = 26.25
 
 ```
+
+# Question 2
+
+```math
+
+h = b \sin A \\
+
+h = 11.3 \sin 32 \\
+
+h = 5.99 \\
+
+\because h \le 6.8 \le 11.3 \\
+
+\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.
+```
+-------------------------
+$`\text{ Case } 1:`$
+
+$`\angle CB^\prime T = \sin^{-1} \Bigl(\dfrac{5.99}{6.8} \Bigr)`$
+
+$`\angle CB^\prime T = 61.75^o`$
+
+$`\angle AB^\prime C = 180 - 61.75 = 118.25^o (\text{ Complentary Angle Theorem})`$
+
+$`\angle ACB^\prime = 180 - 61.75 - 32 = 86.25^o (\text{ASTT})`$
+
+$`\dfrac{AB}{\sin \angle ACB^\prime} = \dfrac{CB^\prime}{\sin A}`$
+
+$`\dfrac{AB}{\sin86.25} = \dfrac{6.8}{\sin 32}`$
+
+$`AB = \dfrac{\sin 86.25 \times 6.8}{\sin32}`$
+
+$`AB = 12.8`$
+
+-------------------------
+$`\text{ Case} 2: `$
+
+$`\angle ABC = 61.75`$
+
+$`\angle ACB = 180 - 32 - 61.75 = 86.25^o (\text{ ASTT})`$
+
+$`\dfrac{AB}{\sin C} = \dfrac{CB}{\sin A}`$
+
+$`\dfrac{AB}{\sin 86.25} = \dfrac{6.8}{\sin 32}`$
+
+$`AB =