From adf0d364898650994aa5db20494a33d9a9d1a4f4 Mon Sep 17 00:00:00 2001
From: James Su <amagicalsoup@gmail.com>
Date: Mon, 30 Dec 2019 03:12:05 +0000
Subject: [PATCH] Add new file

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+## 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)