From 91d06419d72d707940394356c64f04372b7e0978 Mon Sep 17 00:00:00 2001 From: James Su Date: Mon, 25 Nov 2019 21:26:56 +0000 Subject: [PATCH] Update Unit 4: Trigonometry.md --- Grade 10/Math/MPM2DZ/Unit 4: Trigonometry.md | 101 ++++++++++++++++++- 1 file changed, 100 insertions(+), 1 deletion(-) diff --git a/Grade 10/Math/MPM2DZ/Unit 4: Trigonometry.md b/Grade 10/Math/MPM2DZ/Unit 4: Trigonometry.md index 825dfc9..695d464 100644 --- a/Grade 10/Math/MPM2DZ/Unit 4: Trigonometry.md +++ b/Grade 10/Math/MPM2DZ/Unit 4: Trigonometry.md @@ -1,3 +1,5 @@ +# Unit 4: Trigonometry + ## Angle Theorems 1. ```Transversal Parallel Line Theorems``` (TPT) @@ -55,12 +57,18 @@ If three sides of a triangle are respectively equal to the three sides of another triangle, then the triangles are congruent + + ### Side-Angle-Side (SAS) If two sides and the **contained** angle of a triangle are respectively equal to two sides and the **contained** angle of another triangle, then the triangles are congruent. + + ### Angle-Side-Angle (ASA) If two angles and the **contained** side of a triangle are respectively equal to two angles and the **contained** side of another triangle, then the triangles are congruent. + + ## Similary Triangles `Similar`: Same shape but different sizes (one is an enlargement of the other) @@ -83,10 +91,101 @@ our big triangle's area is equal to $`\dfrac{k^2bh}{2}`$. Similar equations and Three pairs of corresponding sides are in the **same ratio** + + ### Side Angle Side similarity (RAR $`\sim`$) Two pairs of corresponding sides are proportional and the **contained** angle are equal. + + ### Angle-Angle similarity (AA $`\sim`$) -Two pairs of corresponding angles are equal. +Two pairs of corresponding angles are equal. In the diagram below, we can solve for the missing angle using Angle Sum Of A Triangle Theorem (ASTT) and see that those 2 triangle's angles are equal. + + + + + +## Primary Trigonometry Ratios + +|Part Of Triangle|Property| +|:---------------|:-------| +|Hypotenuse|The longest side of the right triangle. it is across the $`90^o`$ (right angle)| +|Opposite|The side opposite to the reference angle| +|Adjacent|The side next to the reference agnle| + +**Remember**: Primary Trigonometry ratios are only used to find the **acute** angles or sides of a **right-angled** triangle + +### SOH CAH TOA + +**SINE** $`\sin \theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}`$ + +**COSINE** $`\cos \theta = \dfrac{\text{Adajacent}}{\text{Hypotenuse}}`$ + +**TANGENT** $`\tan \theta = \dfrac{\text{Opposite}}{\text{Adajacent}}`$ + +## Angle Of Elevation And Depression + +| |Angle of Elevation|Angle of Depression| +|:---------|:-----------------|:------------------| +|Definition|**Angle of Elevation** is the angle from the horizontal looking **up** to some object|**Angle of Depression** is the angle frorm the horizontal looking **down** to some object| +|Diagram||| + + +We can see that **Angle of Elevation = Angle of Depression** in the diagram below (Proven using Z-pattern) + + + +## Sine Law + +In any $`\triangle ABC`$: $`\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}`$ or $`\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}`$ + +We can derive the formula further to get: +- $`\dfrac{\sin A}{\sin B} = \dfrac{a}{b}`$ +- $`\dfrac{\sin A}{\sin C} = \dfrac{a}{c}`$ +- $`\dfrac{\sin B}{\sin C} = \dfrac{b}{c}`$ + +Also, for some trigonometry identities: +- $`\tan x = \dfrac{\sin x}{\cos x}`$ +- $`\sin^2 A + \cos^2 A = 1`$ + +**If you are finding the sides or agnles of an `oblique triangle` given 1 side, its opposite angle and one other side or angle, use the sine law.** + +### Ambiguous Case +The ambiguous case arises in the SSA or (ASS) case of an triangle, when you are given angle side side. The sine law calculation may need to 0, 1, or 2 solutions. + +In the ambigouous case, if $`\angle A, a, b`$ are given, the height of the triangle is $`h= b\sin A`$ + + +|Case|If $`\angle A`$ is **acute**|Condition|# & Type of triangles possible| +|:---|:---------------------------|:--------|:-----------------------------| +|1 ||$`a \lt h`$|no triangle exists| +|2 ||$`a = h`$|one triangle exists| +|3 ||$`h \lt a \lt b`$|two triangle exist (one acute triangle, one obtuse triangle)| +|4 ||$`a \ge b`$|one triangle exists| + +|Case|If $`\angle A`$ is **obtuse**|Condition|# & Type of triangles possible| +|:---|:----------------------------|:--------|:-----------------------------| +|5 ||$`a \le b`$|no triangle exists| +|6 ||$`a \gt b`$|one triangle exists| + + +## Cosine Law + +In any $`\triangle ABC`$, $`c^2 = a^2 + b^2 - 2ab\cose C`$ + +**If you are given 3 sides or 2 sides and the contained angle of an `oblique triangle`, then use the consine law** + +## Directions + +`Bearins`: **Always** start from **North**, and goes **clockwise** +`Direction`: Start from the first letter (N, E, S, W), and go that many degrees to the second letter (N, E, S, W) + +**Note:** Northeast, Southeast, NorthWest etc. all have 45 degrees to the left or the right from their starting degree (0, 90, 180, 270) + +## 2D Problems +**Note:** Watch out for the case where you don't know which side the 2 things (buildings, boats etc) are, they can result in 2 answers + +## 3D problems +**Note:** Use angle theorems to find bearing/direction angle, and to help with the problem in general. Apply sine law, cosine law, and primary trigonometry ratios whenever necessary.