Question 1


\because \angle B^\prime = \angle B \quad (\text{Corresonding 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 \dfrac{AB^\prime}{B^\prime C^\prime} = \dfrac{AB}{BC} \\ 

\quad \\ 


\therefore \dfrac{30}{14} = \dfrac{30+x}{22} \\ \\

\quad \\ 

14(30+x) = 22(30) \\

\quad \\

x = \dfrac{22(30)}{14} - 30 \\

\quad \\

x = 17.1428571 \approx 17.14


\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

Question 2


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 =

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