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Grade 10/Math/MPM2DZ/Trig Quiz 1.md
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# 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 =
$`AB = \dfrac{\sin 86.25 \times 6.8}{\sin 32}`$
$`AB = 12.8`$
# Question 3
$`\text{let the square be } ABCD \text{ and the inner triangle } AEF `$
$`\sin (\beta) = \dfrac{EF}{AE} = \dfrac{EF}{1} = EF`$
$`\sin (\alpha) \sin(\beta) = \dfrac{EF}{AE} \times \dfrac{EC}{EF} = \dfrac{EC}{AE} = \dfrac{EC}{1} = EC`$
$`\cos(\alpha) \sin(\beta) = \dfrac{CF}{EF} \times \dfrac{EF}{AE} = \dfrac{CF}{AE} = \dfrac{CF}{1} = CF`$
$`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \sin(\alpha + \beta) = \dfrac{CD}{AE} = \dfrac{CD}{1} = CD`$
$`\cos(\alpha) \cos(\beta) = \dfrac{AD}{AF} \times \dfrac{AF}{AE} = \dfrac{AD}{1} = AD`$
$`\sin(\alpha) \cos(\beta) = \dfrac{FD}{AF} \times \dfrac{AF}{AE} = \dfrac{FD}{1} = FD`$
$`\cos(\beta) = \dfrac{AF}{AE} = \dfrac{AF}{1} = AF`$
$`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \cos(\alpha + \beta) = \dfrac{BE}{AE} = \dfrac{BE}{1} = BE`$