3.3 KiB
Question 1
\(`\because \angle B^\prime = \angle B \quad (\text{PLT-F})`\)
\(`\because \angle C^\prime = \angle C \quad (\text{PLT-F})`\)
\(`\therefore \triangle AB^\prime C^\prime \sim \triangle ABC \quad (\text{AA } \sim) `\)
\(`\therefore \dfrac{AB^\prime}{B^\prime C^\prime} = \dfrac{AB}{BC} `\)
\(`\therefore \dfrac{30}{14} = \dfrac{30+x}{22}`\)
\(`14(30+x) = 22(30) `\)
\(`x = \dfrac{22(30)}{14} - 30 `\)
\(`x = 17.1428571 \approx 17.14 `\)
\(`\therefore \dfrac{AC^\prime}{B^\prime C^\prime} = \dfrac{AC}{BC} `\)
\(`\therefore \dfrac{y}{14} = \dfrac{y+15}{22} `\)
\(`22y = 14y + 14(15) `\)
\(`8y = 14(15) `\)
\(`y = 26.25`\)
Question 2
\(`h = b \sin A`\)
\(`h = 11.3 \sin 32`\)
\(`h = 5.99`\)
\(`\because h \lt 6.8 \lt 11.3`\)
\(`\therefore 2 \triangle 's \text{ exist}`\)
\(`\text{ Lets call point } T \text{ is the height that is perpendicular on side } AB \text{ and connects to point } C. \text { and } B^\prime \text{ be the other possible point of } B.`\)
| \(`\text{ Case } 1:`\) |
|---|
| \(`\text{ Case } 2: `\) |
| \(`\angle ABC = \angle CB^\prime T = 61.75^o (\text{ITT})`\) |
| \(`\angle ACB = 180 - 32 - 61.75 = 86.25^o (\text{ASTT})`\) |
| \(`\dfrac{AB}{\sin C} = \dfrac{CB}{\sin A}`\) |
| \(`\dfrac{AB}{\sin 86.25} = \dfrac{6.8}{\sin 32}`\) |
| \(`AB = \dfrac{\sin 86.25 \times 6.8}{\sin 32}`\) |
| \(`AB = 12.8`\) |
\(`\therefore AB \text{ could either be } 6.37cm \text{ or } 12.8cm`\)
Question 3
\(`\text{let the square be } \square ABCD \text{ and the inner triangle be } \triangle AEF `\)
\(`\sin (\beta) = \dfrac{EF}{AE} = \dfrac{EF}{1} = EF`\)
\(`\sin (\alpha) \sin(\beta) = \dfrac{EC}{EF} \times \dfrac{EF}{AE} = \dfrac{EC}{AE} = \dfrac{EC}{1} = EC`\)
\(`\cos(\alpha) \sin(\beta) = \dfrac{CF}{EF} \times \dfrac{EF}{AE} = \dfrac{CF}{AE} = \dfrac{CF}{1} = CF`\)
\(`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \quad \sin(\alpha + \beta) = \dfrac{CD}{AE} = \dfrac{CD}{1} = CD`\)
\(`\cos(\alpha) \cos(\beta) = \dfrac{AD}{AF} \times \dfrac{AF}{AE} = \dfrac{AD}{1} = AD`\)
\(`\sin(\alpha) \cos(\beta) = \dfrac{FD}{AF} \times \dfrac{AF}{AE} = \dfrac{FD}{1} = FD`\)
\(`\cos(\beta) = \dfrac{AF}{AE} = \dfrac{AF}{1} = AF`\)
\(`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \quad \cos(\alpha + \beta) = \dfrac{BE}{AE} = \dfrac{BE}{1} = BE`\)
\(`\therefore \sin(\alpha + \beta) = CD = CF + FD = \cos(\alpha) \sin(\beta) + \sin(\alpha) \cos(\beta)`\)
\(`\therefore \cos(\alpha + \beta) = BE = \cos(\alpha - \beta)`\) ## Footer Notes
\(`\text{ASTT} = \text{ Angle Sum Of Triangle Theorem}`\)
\(`\text{ITT} = \text{ Isoceles Triangle Theorem}`\)
\(`\text{SAT} = \text{ Corresponding Angle Theorem}`\)
\(`\text{PLT-F} = \text{ Parallel Line Theorem; F-pattern}`\)
\(`\text{AA} \sim \space = \text{ Angle Angle Similarity Theorem}`\)