ID#3: Unit Q Concept 1

Where does sin2x+cos2x=1 come from?
Thinking back to the unit circle, this rule comes from the pythagorean theorem which states "a2+b2=c2".The pythagorean theorem can be written to be x^2+y^2=r^2, this is possible because when we look at a triangle from the unit circle two of its legs are x and y and the hypotenuse is r. But again, when we think back to the unit circle, we want the equation to equal 1, so we divide everything by r to get (x/r)^2 + (y/r)^2 = 1. because we have our identities memorized, we realized that (x/r) and (y/r) have corresponding trig functions. With that being said, we know that Cos= x/r and Sin= y/r. when we plug that back into the equation, we are left with sin2x+cos2x=1. that is why this formula is referred to as the pythagorean identity.

WPP #13 & 14: Unit P Concepts 6 & 7

This WPP was made in collaboration with Joe Castillo. Please visit the other awesome posts on their blog by going here


Anthony Dufegt has survived an extremely hot day so far today and decides he's had enough of this constant struggle against the heat alone, so he decides to go to the pool. He walks out of his house and turns N 30° W then turns 100° due north, after walking for 60 meters more, he finally arrives at the pool. But alas, before he can get in, he ponders how many meters he would've walked if he could've just walked in a straight line instead. After he has finished his "underwater aerobics" class, he decides to take a different route and walks east 30 meters then turns at a bearing of 150° and continues to walk for another 80 meters until he gets home. Again, as he arrives home, he wonders how much he would've walked if had, instead, walked in a straight line. 

Part 1-Law of Sines (going to the pool)

Part 2-Law of Cosines (coming back from the pool)

BQ #1: Unit P Concepts 1 & 4: Law of Sines and Area Formulas

1. Law of Sines: Why do we need it?  How is it derived from what we already know?
We use the law of sines to solve for non-right triangles.This formula helps us find missing sides or angles when we only have AAS (angle, angle, side), ASA (angle side angle), or SSA (side, side, angle). We use the trigonometric function of sine to derive the formula and form two right triangles then we are able to derive for the formula. But before we do, we will have to add other variables to make it possible, like 'h' for the height (perpendicular to the base). If we draw a perpendicular line, we will get our 2 right triangles, which we need to derive for the formula. The following pictures go over the derivation of the formula.

4. Area Formulas: How is the “area of an oblique” triangle derived?  How does it relate to the area formula that you are familiar with?
The area of an oblique triangle is derived from the area of a triangle formula: A=1/2 bh. This formula allows us to find the area of a triangle without having to know what the value of "h" is. The formula for an oblique triangle is similar to the formula of a regular triangle because they both have the same purpose and the same root or origin. The only difference is that in the original formula we must know the value of "h", but in the oblique triangle formula we don't need to know the value of "h" because instead we will be using one of the angle measures. The picture below will go over the derivation of the formula.

WPP #12: Unit O Concept 10: solving angle of elevation and depression word problems

Introduction: The PETCO across from Jose's bakery. 

Problem: The PETCO across from Jose's bakery

Jose is standing atop his bakery looking out towards the PETCO across the street that he often visits to buy neat things for his dog Simba. The street between the two buildings is 12 meters long. He looks down at the entrance to see a customer walking in with a dog that reminds him of Simba. He looks down at an angle of depression of 38°, this makes him wonder how tall PETCO is from his level of elevation. He then looks up at an angle of elevation of 34°, staring directly at the top of PETCO. How tall is PETCO from the height of his building and what is the total height of PETCO?

Solution: The PETCO from across Jose's bakery

PART 1 ANSWER: First we solve for how tall PETCO is from his level of elevation: tan38=x/12. in order to solve for x, we must multiply the entire problem by 12 then solve, which equals 8.09.

PART 2 ANSWER: then we solve for the rest of the building's height: tan34=x/12. to solve for x, we must also multiply the entire problem by 12 then solve, equaling 9.38. 

So to get the total height of PETCO, we must add the two distances of 8.09 meters and 9.38 meters, equaling 17.47 meters.

I/D2: Unit O- How can we derive the patterns for our special right triangles?

Inquiry Activity Summary

30-60-90 Triangles

Before we can solve a 30-60-90 triangle, we need to know how to get 30-60-90 triangles. On this worksheet we are given a 60-60-60 triangle (equiangular), and in order to get a 30-60-90, we must cut down the middle of it, giving us 2 triangles, both 30-60-90. To derive the patterns for 30-60-90 triangles, we must first know what to use to solve for each side. One way to get the side measures is using a special pattern. This pattern says that the shortest side is 'n', the hypotenuse is '2n', and the height is 'n√3'. But because this triangle is a right triangle, we can also use the Pythagorean theorem. But we need two sides in order to solve for one missing side; unfortunately we are only given one side measure: the hypotenuse (equals 1), that isn't an issue in this case because in these specific triangles, the shortest side is always half the size of the hypotenuse. Now we have the size of 2 sides; now we can use the Pythagorean theorem: a²+b²=c² . For this example, I'll label the short leg "b".  a²+(⅟₂)² =1² so simplified, a= ^√3/_2. 

45-45-90 Triangles

Deriving the pattern for a 45-45-90 triangle is simple and somewhat similar to the previous triangle. In order to get a 45-45-90 triangle, we must split the square in half diagonally, we do so because each angle in a square is 90˚ and splitting the opposites in half gives us 45˚ on both sides; leaving us with a 45-45-90 triangle. Because this triangle is also a right triangle, we can also use the Pythagorean theorem to solve for missing sides. In this activity we have the side lengths. For any of these triangles, "a" and "b" are the same lengths, and because of this fact about 45-45-90 triangles, filling in the Pythagorean formula is easier. 1²+1²=c² is the result of substituting in the formula, fully simplified: c=√2.

INQUIRY ACTIVITY REFLECTION

Something I never noticed about special right triangles is... that the easiest way to solve for sides is to use the Pythagorean theorem.

Being able to derive these patterns myself aids in my learning because... it gave me a visual representation which helped me better my understanding of special right triangles.