SP #7: Unit Q Concept 2 (Finding All Trig Functions When Given One Trig Function and Quadrant)

Solving with SOH CAH TOA:

1.

2.Solving with Identities:

When finding actual trig functions, we can use two methods: SOH-CAH-TOA or Identities! As you can see from the first photo, we refer back to our Unit circle and can use the Pythagorean Theorem to get our missing side with the two we already have. In our case, we go to the sin where we know the fraction is y/r. If we place it into the formula, our "x" will be one! The second method is through identities! When using our identities, we are always given two trigs which cuts down a portion of our work! With these two, we first figure out in which quadrant the answers will be either negative or positive in (please refer to the second picture). After that is when we either substitute or solve for the proper trig that we are looking for.

BQ #1: Unit P Concepts 6-7 (Law of Sines and Law of Cosines)

Law of Sines

1. Why do we need it?

• First and foremost, we have to understand that not all triangles are right triangles. This means that not all triangles have a ninety degree angle, meaning we cannot use the Pythagorean theorem to solve for more. Then how will we solve  for irregular triangles you ask? Fear not, my beloved friend because this is where the Law of Sines come in to play.

-The Law of Sines, however, may only be used when given SSA (side-side-angle)

2. How do we derive from what we already know?

• From what we already know, the triangle does not have a right triangle. this is where we draw a Line straight down the middle to make two right triangles. This length is labeled as "h."

 

http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

 

• If we look at this from a graph perspective, we see that the sin<A will equal h/c because it's opposite over adjacent. If we multiply both sides by "c" then we would get cSin<A=h. Sin<C will equal h/a and multiply by "a" then aSin<C will also equal "h" which will make all angles of the same value.
• This can also be said for angle B therefore we plug Sin<A/value = Sin<B/x and cross multiply! This is how we derive the Law of Sine.

 

I/D #2: Unit O Concepts 7-8

Inquiry Summary Activity:

A. 30-60-90 triangle
 We can derive a special right triangle of 30-60-90 angles from an equilateral triangle. An equilateral triangle is a triangle that has all three angles of the same degree to add up to 360 degrees which would make them 60 degrees. Along with having all angles of same degree, all sides are of equal length as well. In this case one. We derive this special right triangle by cutting the equilateral. By doing this we create two triangles of 30-60-90 degrees, however, we just need to use one.

Because one side was unaffected by the split, we know that it has a side length of one while we know that the other original side was cut in half giving us 1/2. With these two side lengths, we can use the Pythagorean Theorem to get the missing length of b- theorem being a^2 + b^2 = c^2. By solving, we get the value of radical3/2.

 

After every value has been found, we place them in their sides such as the photo below. In looking at the picture, we can see that fractions can be bothersome. A method to remove fractions would be to multiply all sides by two to become:

 

 

The original values can be labeled as "n" or really any variable because they represent the basis of the Special Right Triangle (SRT). Here is when we begin to see the pattern completely because no matter what the length values will be, n will always be connected to the 30-60-90.

 

 

B. 45-45-90 Triangle

 



The 45-45-90 triangle is much like the one in prior concept of 30-60-90. The difference is, we derive these from a square. Our lengths are already given to us to have an value of one all the way around. Again we are going use a line to cut the shape differently. Please refer to the photo below.

 

 

This would have cut two of the 90 degree angles to make four forty-five degree angles, giving us once again two new angles. This time, our degrees would be a special right triangle of 45-45-90 degrees. Because the outer lengths, also known as the legs (or a and b) were unaffected by the cut, we can once again use the Pythagorean theorem to get the missing leg of the hypotenuse. Your work should be plugged in and solved similarly to the photo below.

 

 

 

Because our leg lengths are once again the value of one, we want to replace it with variable "n." This "n" now represents the basic pattern value of a 45-45-90 triangle because it applies to any length of this type of triangle.  This will give us two legs with the value of n and our hypotenuse with n times radical two. Keep in mind that if the value of the triangle is expanded, it must be multiplied by the original values of n and so on.

 

 

 

Inquiry Activity Reflection:

1. “Something I never noticed before about special right triangles is…” That there is always a pattern in finding values of sides.

2. “Being able to derive these patterns myself aids in my learning because…” they ensure that I will remember the values or the proper method of getting any of my missing values.