Unit Circle

Hello guys, this is Luis posting. Mr. K started the class by giving us some advice/tips on how to properly use the internet as well as internet etiquette and I myself was really amazed on what I had heard. We discussed the homework assignment that he gave us regarding Degree and Radian, converting Degree and Radian and vice versa. Some questions in the assignment were just a review that we encountered in Pre-cal 11. Most of us struggled to solve the questions in the assignment because we almost forgot how to do it.

You can quickly look at the problems on the above slides and these are their brief explanations. If you have any questions on how we came up with the answer, feel free to ask me or Mr. K.

Oliver has already posted the conversion of Degrees to Radian and vice versa and this first slide is just another example of it.

The 3rd slide is question number 5 Exercise 1 on our assignment, we all know that π is equal to 180. So, if the measure of the whole triangle is equal to 180 it therefore equals to π. Mr. K. told us that we can solve this through mental math. We just have to add the two given angles and subtract it from π.

On the 4th and 5th slides, he showed us the easier way to solve parallel lines intersected by a transversal. In his example the corresponding angles have the same size and shape therefore those angles are equal. We used the formula D/2π=R/2πr.

He also gave a brief recap of what we had learned in Pre-Cal 11 such as Laws of Exponents, simplifying radicals and using quadratic formula.





On the 13th slide Mr. K. gave us a problem to solve and one of our classmates came up with the correct answering using the formula S= θr. Although, she came up with the correct answer, S= θr can only be used in one circumstance which may be the reason why Mr. K doesn’t teach this formula. The formula D/360=L/2πr can be used in many places, though it’s a long process it is more concise and more proportional.


On the other hand, if the given is asking for Radian, instead of using D/360=L/2πr we can use the 2π which is equal to 360 to express the answer. The formula will be D/2π=L/2πr.

For example:
Determine the length of the arc of the circle with the radius 8 cm that subtends each angle at the centre. Express each length to one decimal place if necessary.

1. 30°

Given:
Radius= 8
Degree = 30°

D/360=L/2πr
30°/360=L/2π(8)

0 = 30°(16π)/360
0 = 4.2°


On the 18th to 20th slides, we had used the SOHCAHTOA which means:

SOH used for sin which is sin θ= opposite / hypotenuse
CAH used for cos which is cos θ = adjacent / hypotenuse and
TOA used for tangent θ = opposite / adjacent

In this triangle the hypotenuse is 40. Why 40? Because the hypotenuse will always be twice the size of the opposite side, and the opposite side will always be half the size of hypotenuse.

To find the adjacent, we can use the cosine rule or the Pythagorean Theorem.

a^2+b^2 = c^2
b^2 = c^2 – a^2
= 40^2 – 20^2
= root of 1200
= 20 root of 3

On the 20th slide, they are all corresponding angles because they are all equal to 30° even though they are on the different quadrants. So, basically 150° is the same as 30°. The Quadrant I and the Quadrant IV are positive because the adjacent of the two angles are on the positive side of the x-axis. However, the II and III quadrants are both negative because two adjacent are on the negative side of the x-axis.

On the 21st slide, he talked about the reference angle. On his example, sine 300 and sine -60 is both equal to -0.8666. The reference angle of 300 is 60.

On the last slide, he then had given us a short recap on how to simplify radicals.

The next scribe is Precious.