When To Use What Trig?


SOH CAH TOA:  These laws can be used when you have a right triangle, you have one angle, and one side (for the same triangle).

“Opposite, Adjacent and Hypotenuse” refer to the length of the side relative to θ (Theta). To use this formula you must have two of the variables in the equation.

Sin θ = Opposite/Hypotenuse:

Cosine θ = Adjacent/Hypotenuse

Tangent θ = Opposite/Adjacent

triangle showing Opposite, Adjacent and Hypotenuse


Sine Law: Sin A/a = Sin B/ b = Sin C/c: This law is used for looking for a side or an angle that is missing on a non-right triangle.

A, B, C are representative of the angles on a triangle. Any three letters in alphabetical order may be used to represent the angles.

a, b, c are representative of the side lengths of a triangle –the same rule applies to small letters that applied to capitals.

This law can be used to find the Ambiguous Case on a triangle (a triangle in which two side lengths are given and one angle opposite to a side is given). The Ambiguous Case is written as A-S-S most often, “S” means side and “A” means angle.

This law  may only be used when either two side lengths and one opposite angle is given, or when two angles and one opposite side is given.



Cosine Law: a^2= b^2 + c^2 – 2bc * Cos A : This law can be used to find a side or an angle missing on a non-right triangle.

The letter rules that applied to Sine Law apply here.

This rule may be used when two opposite sides and one angle in between is given, or when three of the variables in the equation are given. This law is often written as S-A-S.

The Cos A may be replaced by either Tangent A or Sine A.


trig cos rule example


Reflection on Math 10:

In Math 10 you learned how to use SOH CAH TOA to find the sides of a right triangle, this year it is an expansion on every thing. This year we gained the skills to solve non right triangle. d

Flame Test Lab


Weight Station Mole Lab


The formulae and steps we used to find the weight of 0.126 mol of Sodium Carbonate were:

  1. Find the weight of Na2CO3 (23*2) + 12 + (16 * 3)= 106 g/mol
  2. Multiply the mol by our g/mol (0.126 mol of Na2CO3 *106 g /1mol = 13.30 grams)
  3. Weigh them out and zero out the weigh boat

Using all of these steps we were able to conclude that the weight of 0.126 mol of sodium carbonate is 13.3 grams.




Chem 11: Types Of Reactions -Practical Lab


The video above is a Magnesium Ribbon being set on fire. In around ten seconds the magnesium burned out and emitted a very bright light, as seen in the video.  This type of reaction was a synthesis combining Magnesium with Oxygen.


This video is an example of a combustion. A small butane lighter burns butane (C4H10) and oxygen together (O). The butane caught on fire as soon as the red trigger was clicked.


This reaction showed hydrogen peroxide bubbling up, steaming & changing colour. Potassium Iodide was added to the hydrogen peroxide, but it simply sped up the process that would happen naturally in different circumstances.


copperag5 copperag3copperlast

This series of pictures and videos shows the progression that occurs when copper and silver nitrate are added together and mixed. The silver which was in liquid form at the beginning suddenly becomes visible as the elements react together. The final picture took around 8 1/2 minutes to take. This was the slowest reaction out of all the other 5.


This video shows potassium iodide reacting and turning yellow once lead nitrate is added. This reaction happened very quickly, as quickly as the butane set on fire.






This series of videos shows sodium hydroxide and sulfuric acid reacting together to balance each-other out. The litmus paper changes colours (or stays the same) depending on whether the solution is acidic or basic.


Week 9, Math Burton

What I learned this week:

I learned two interesting things this week, the first one was that an equation that repeats itself with the same numbers is really just a square in disguise. Example: (X +4)(X + 4). Usually you would use FOIL to figure out the answer, redistributing and multiplying every number. OR you can think of it as (X + 4)^2, you can cut one step out as well to find the middle number. Multiply 4*X*2.


Second thing I learned (X + 5) (X + 3) usually you distribute everything, or you can square the X (since you know you’d do that anyways), add the two numbers together (add x) and lastly, multiply the two numbers.

This: X*X + X*3 + 5*X + 5*3


X^2 + 8X + 15