# 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 $`h = b \sin A`$ $`h = 11.3 \sin 32`$ $`h = 5.99`$ $`\because h \lt 6.8 \lt 11.3`$ $`\therefore 2 \triangle 's \text{ exist}`$ $`\text{ Lets call point } T \text{ is the height that is perpendicular on side } AB \text{ and connects to point } C. \text { and } B^\prime \text{ be the other possible point of } B.`$ ------------------------- $`\text{ Case } 1:`$ $`\angle CB^\prime T = \sin^{-1} \Bigl(\dfrac{5.99}{6.8} \Bigr)`$ $`\angle CB^\prime T = 61.75^o`$ $`\angle AB^\prime C = 180 - 61.75 = 118.25^o (\text{CAT})`$ $`\angle ACB^\prime = 180 - 118.25 - 32 = 29.75^o (\text{ASTT})`$ $`\dfrac{AB}{\sin \angle ACB^\prime} = \dfrac{CB^\prime}{\sin A}`$ $`\dfrac{AB}{\sin29.75} = \dfrac{6.8}{\sin 32}`$ $`AB = \dfrac{\sin 29.75 \times 6.8}{\sin32}`$ $`AB = 6.37`$ ------------------------- $`\text{ Case } 2: `$ $`\angle ABC = \angle CB^\prime T = 61.75^o (\text{ITT})`$ $`\angle ACB = 180 - 32 - 61.75 = 86.25^o (\text{ASTT})`$ $`\dfrac{AB}{\sin C} = \dfrac{CB}{\sin A}`$ $`\dfrac{AB}{\sin 86.25} = \dfrac{6.8}{\sin 32}`$ $`AB = \dfrac{\sin 86.25 \times 6.8}{\sin 32}`$ $`AB = 12.8`$ -------------------------- $`\therefore AB \text{ could either be } 6.37cm \text{ or } 12.8cm`$ # Question 3 $`\text{let the square be } \square ABCD \text{ and the inner triangle be } \triangle AEF `$ $`\sin (\beta) = \dfrac{EF}{AE} = \dfrac{EF}{1} = EF`$ $`\sin (\alpha) \sin(\beta) = \dfrac{EC}{EF} \times \dfrac{EF}{AE} = \dfrac{EC}{AE} = \dfrac{EC}{1} = EC`$ $`\cos(\alpha) \sin(\beta) = \dfrac{CF}{EF} \times \dfrac{EF}{AE} = \dfrac{CF}{AE} = \dfrac{CF}{1} = CF`$ $`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \quad \sin(\alpha + \beta) = \dfrac{CD}{AE} = \dfrac{CD}{1} = CD`$ $`\cos(\alpha) \cos(\beta) = \dfrac{AD}{AF} \times \dfrac{AF}{AE} = \dfrac{AD}{1} = AD`$ $`\sin(\alpha) \cos(\beta) = \dfrac{FD}{AF} \times \dfrac{AF}{AE} = \dfrac{FD}{1} = FD`$ $`\cos(\beta) = \dfrac{AF}{AE} = \dfrac{AF}{1} = AF`$ $`\text{Draw a parallel line to } CD \text{ that connects point } E \text{ to } AD. \quad \cos(\alpha + \beta) = \dfrac{BE}{AE} = \dfrac{BE}{1} = BE`$ ## Footer Notes $`\text{ASTT} = \text{ Angle Sum Of Triangle Theorem}`$ $`\text{ITT} = \text{ Isoceles Triangle Theorem}`$ $`\text{CAT} = \text{ Corresponding Angle Theorem}`$ $`\text{PLT-F} = \text{ Parallel Line Theorem; F-pattern}`$ $`\text{AA} \sim \space = \text{ Angle Angle Similarity Theorem}`$