## Thursday, March 20, 2014

### I/D#3: Unit Q Concept 1: Using Fundamental Identites to Simplify & Verify Expressions

Inquiry Activity Summary

"sin²x+cos²x=1" comes from the identity (proven factor or formula) of the Pythagorean Theorem. The theorem, when involving the trig ratios of the Unit Circle, is written in the form of "x²+y²=r²". In order to get the set of values to equal "1", we must divide the entire equation by "r²". As a result, we are left with two ratios, "x/r" (cosine), and "y/r" (sine).

Inquiry Activity Reflection

The connections that I see between Units N, O, P, and Q so far are the use of angels in degrees and the use of geometric properties. Likewise, I have noticed that all of the units make use of the six central trigonometric ratios.

If I had to describe trigonometry in THREE words, they would be: patterns, formulas, and memorization.

## Tuesday, March 18, 2014

### WPP #13 & 14: Unit P Concepts 6 & 7: Applications With Law of Sines and Law of Cosines

This WPP #13 & 14 was made in collaboration with my friend and classmate, Jennifer G. Please visit her blog here to view her other amazing blog posts!

## Monday, March 17, 2014

### BQ #1: Unit P Concepts 2, 4 & 5: Law of Sines (SSA) & Area Formulas

2. Law of Sines
Why is SSA ambiguous? Accurately draw the triangles that would be associated with one of these problems (pick any from the SSS or PQ). Connect your answer to your knowledge of Unit Circle trig function values.
SSA is ambiguous because we are only given one of the three angles as opposed to two in ASA or AAS. Because of this, there is the chance that we may get an answer of one, two, or no possible triangles.

5. Area Formulas

Show that our “traditional” area formula of A=1/2bh, the area of an oblique triangle, and Heron’s area formula will result in the same values. Draw and label a right triangle with angles of 35* and 65*. Have the base be 4 units. Find all the remaining pieces using your knowledge of this chapter and then find the area using all three area formulas.

Our "traditional" area formula, the area of an oblique triangle, and Heron's area formula will all result in the same value for the area of a triangle.

## Tuesday, March 4, 2014

### I/D#2: Unit O: How Can We Derive the Patterns for Special Right Triangles?

Inquiry Activity Summary

The objective of the activity below was to derive or obtain the patterns of special right triangles from equilateral shapes, in this case a larger triangle and a square. The benefits of completing this task is that it provides us with another means of identifying the Unit Circle values.

 Activity page 1 of 2. Click to enlarge.
 Activity page 2 of 2. Click to enlarge.

## 45°, 45°, 90° Triangle

As shown in the Step 1 of the above activity (see page 1), the process of deriving a 45°, °45, 90° congruent triangles of the desired angles. We label the hypotenuse of the smaller triangles "c", and we can label the two legs "a" and "b". Based on what we informed in this activity, we know that the side lengths of the square were valued at 1; thus, both "a" and "b" are equal to 1. To find the value of the hypotenuse, we can use the Pythagorean Theorem (a² + b² = c²). By plugging in the corresponding values, we get c= √2. We can then collectively use the newly-obtained values and a variable (e.g. "n") to triangle from a square first involves the act of dividing, or breaking down, the initial shape. To do so, we slice the square diagonally to from two form a general pattern. This means that by adding a variable to each of the values, the pattern becomes applicable to a square with any given side length!

## 30°, 60°, 90° Triangle

In Step 2 of the activity (see page 2), we are presented with an equilateral triangle with a side length of 1. Referring back to what we have learned in Geometry, we know that a triangle with sides of equal length is also equiangular, meaning that each of the three angles are 60°. To begin this step, me must again breakdown the shape by dividing it into congruent parts--this time by drawing a vertical line through the middle. Again we label out hypotenuse "c" and the two legs of the shape "a" and "b". Using the given information, we know that "a" is equal to 1/2 and "c" is equal to 1. As you may have probably already guessed from Step 1, we use the Pythagorean Theorem to find the missing height. By plugging in the respective values into the equation and solving for "b", we get the answer of √3 /2. In order to translate our values into the normal pattern without fractions, we multiply each of them by 2. By doing so, "a" becomes 1, "b" becomes √3, and "c" becomes 2. As stated before, we can include a variable (e.g. "n") to our answers to form a general pattern. This allows the pattern to be applicable to an equilateral triangle with any given side length!

## Inquiry Activity Reflection

Something I never noticed before about special right triangle is that they can be joined together to form larger, equilateral and equiangular shapes!

Being able to derive these patterns myself aids my learning because it enables me to identify patterns that could prove useful in future practice and exams.