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95 lines
1.7 KiB
Markdown
95 lines
1.7 KiB
Markdown
# Question 1
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```math
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\because \angle B^\prime = \angle B \quad (\text{Corresonding Line theorem}) \\
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\because \angle C^\prime = \angle C \quad (\text{Corresponding Line theorem}) \\
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\therefore \triangle AB^\prime C^\prime \sim \triangle ABC \quad (\text{ AA } \sim) \\ \\
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\therefore \dfrac{AB^\prime}{B^\prime C^\prime} = \dfrac{AB}{BC} \\
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\quad \\
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\therefore \dfrac{30}{14} = \dfrac{30+x}{22} \\ \\
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\quad \\
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14(30+x) = 22(30) \\
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\quad \\
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x = \dfrac{22(30)}{14} - 30 \\
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\quad \\
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x = 17.1428571 \approx 17.14
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```
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```math
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\dfrac{AC^\prime}{B^\prime C^\prime} = \dfrac{AC}{BC} \\
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\quad \\
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\dfrac{y}{14} = \dfrac{y+15}{22} \\
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\quad \\
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22y = 14y + 14(15) \\
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8y = 14(15) \\
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y = 26.25
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```
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# Question 2
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```math
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h = b \sin A \\
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h = 11.3 \sin 32 \\
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h = 5.99 \\
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\because h \le 6.8 \le 11.3 \\
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\therefore 2 \triangle 's \text{ exist} \\
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\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.
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```
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-------------------------
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$`\text{ Case } 1:`$
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$`\angle CB^\prime T = \sin^{-1} \Bigl(\dfrac{5.99}{6.8} \Bigr)`$
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$`\angle CB^\prime T = 61.75^o`$
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$`\angle AB^\prime C = 180 - 61.75 = 118.25^o (\text{ Complentary Angle Theorem})`$
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$`\angle ACB^\prime = 180 - 61.75 - 32 = 86.25^o (\text{ASTT})`$
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$`\dfrac{AB}{\sin \angle ACB^\prime} = \dfrac{CB^\prime}{\sin A}`$
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$`\dfrac{AB}{\sin86.25} = \dfrac{6.8}{\sin 32}`$
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$`AB = \dfrac{\sin 86.25 \times 6.8}{\sin32}`$
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$`AB = 12.8`$
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-------------------------
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$`\text{ Case} 2: `$
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$`\angle ABC = 61.75`$
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$`\angle ACB = 180 - 32 - 61.75 = 86.25^o (\text{ ASTT})`$
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$`\dfrac{AB}{\sin C} = \dfrac{CB}{\sin A}`$
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$`\dfrac{AB}{\sin 86.25} = \dfrac{6.8}{\sin 32}`$
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$`AB = |