## Trigonometry ### Question 1 a) It means to solve all missing/unknown angles and sidelengths. It can be achieved by using some of the following: 1. Sine/cosine law 2. Primary Trigonometry Ratios 3. Similar / Congruent Triangle theorems 4. Angle Theorems 5. Pythagorean Theorem ### Question 1 b) Draw a line bisector perpendiculr to $`\overline{XZ}`$. Then by using pythagorean theorem: $`7^2 - y^2 = h^2`$, where $`h`$ is the height. $`\therefore h = \sqrt{13} \approx 3.61cm`$ ### Question 1 c) We can draw a triangle $`ABC`$ where $`\angle A`$ is the angle between the hands, and $\overline{AB}`$ and $`\overline{BC}`$ are the long and short hands respectively. Since a clock is a circle, $`\angle A = \dfrac{360}{12} \times 2 = 60^o`$ Let $`x`$ be the distance between the 2 hands. By using the law of cosines: $`x^2 = 12^2 + 15^2 - 2(15)(12)\cos60`$ $`x = 13.7cm`$. The distance between the 2 hands is $`13.7cm`$. ### Question 2 a) Lets split the tree into the 2 triangles shown on the diagram. By using the primary trigonmetry ratios, we know that the bottom triangle's height side lenghts that is part of the tree's height is $`100\tan 10`$, and $`100 \tan 25`$ for the top triangle. Therefore the tree's height is the sum of these 2 triangle's side length. Therefore the total height is $`100(\tan 25 + \tan 10) = 64.3`$ The height of the tree is $`64.3m`$ ### Question 2 b) $`\angle G = 180 - 35 - 68 = 77 (ASTT) `$ By using the law of sines. $`\overline{RG} = \dfrac{173.2 \sin 35}{\sin 77} = 102m`$ By using the law of sines. $`\overline{TG} = \dfrac{173.2 \sin 68}{\sin 77} = 164.8m`$ $`P = 173.2 + 164.8 + 102 = 440m`$ The perimeter is $`440m`$ ### Question 2 c) We know the buildings must be on the same side because they both cast a shadow from the same one sun. Let the triangle formed by the flagpole be $`\triangle FPS`$ and the one by the building $`\triangle TBS`$ $`\because \angle B = \angle P`$ (given) $`\because \angle S`$ is common. $`\therefore \triangle TBS \sim \triangle FPS`$ (AA similarity theorem) $`\dfrac{TB}{BS} = \dfrac{FP}{PS} \implies \dfrac{TB}{26} = \dfrac{25}{10]`$ $`\therefore TB = \dfrac{25(26)}{10} = 65`$ Therefore the building is $`65m`$ tall. ### Question 3 a)