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December 6, 2005   Lesson Review E-34 (Dedicated to Ms. Woodlief, math teacher at Drew Magnet School in the City of Los Angeles)  Enter Chelsea (13) from Drew Magnet School, and sister, Kaylan..Also:  Ke'Ontra returns for Geometry One and Two (below)

Chelsea and I worked problems in Algebra One for over 2 hours. Kaylan entertained herself with one of our computers.

Chelsea is a very mannerable and special young lady. She left me a note on our dry erase board that said "Bye and thanx for the help".

I have tutored and toiled with many students who have never even acknowledged that I did anything to help them! Thank you, Chelsea. Kind words are all that I need to work myself to the bone for you.

We worked with straight lines. I reinforced everything that Chelsea's math teacher, Ms. Woodlief, had taught. I also provided other ways to find the slope (M) and ways to graph the lines. Chelsea had moved ahead of her class, now. I did not realize she had not been instructed in graphing lines, just writing down the slope from ax +by =c. ...where the slope is defined by   - a/b.

Chelsea also put her homework equations in the y= mx + b format for a straight line. She knows that the slope is also defined by (m), which is always a fraction. From (b), which is the point where the line crosses the "Y" axis, one can simply go from that point with the slope, which is y/x. As an example, take the line, y= 3x + 5, the slope is 3/1 and the point at where the line crosses the "y" axis is (0, 5). From (0,5), go 3 divisions (y) up and 1 division over (x). Then make your second point; draw your line through the 2 points and you are finished. The whole process should take no more than 20 seconds!

Chelsea worked many problems. She worked them all correctly!

Earlier, Ke'Ontra (former student) and I worked with Geometry One and Two. I reminded Ke'Ontra to look for the shortest solution to a problem. Case in point; we had a problem that involved finding the area of an equilateral triangle with a height of 9 inches. The solution is a very simple one if it is remembered that a triangle has 180 degrees in it! Each angle then, is 60 degrees. The height splits one of the 60 degree angles to make two 30 degree angles. That is precisely what we want; a right triangle of 30 degrees because we know the cosine of 30 degrees to be "the adjacent side over the hypotenuse: with the adjacent side known to be 9 inches, it is a simple matter to look up the cosine of 30 degrees on the calculator ( √3/2 ), which is 1.732 ÷ 2 = 0 .866 and set it equal to 9/R.

cosine 30 = 9/R: R = 9/cosine 30 or 9/0.866  or 10.39

Then one of the sides of the triangle becomes known. Next use the formula for the area of a triangle; A =½ BH.  Solve the area for yourself.....A= ½ (10.39)(9) = 46.76 inches

One thing leads into another. Having reviewed right triangles, I thought it best to talk about a 30 and 45 degree right triangle; the sine, cosine and tangent of each and exactly what all of that meant. We reviewed opposite side, adjacent side and hypotenuse. We reviewed the Pythagorean Theorem and reviewed how to find an unknown side when two are given in a right triangle.  We looked up the sine, cosine and tangent of 30 and 45 degree angles and reviewed the ratios defined by each. Take a look at a 45 degree right triangle. The tangent is (1) according to the calculator. That means that the (y/x) is 1/1, or the opposite side over the adjacent side is 1/1. We then solve for the unknown side, (R) or the hypotenuse. It is R^2 = A^2 = B ^2......R^2 = (1) ^2 + (1)^2 which is R^2 =2. That makes (R) = √2  No that is precisely what we want to know...the relationship between all sides in a 45 degree right triangle...1, 1, √2  (whatever the over (x) and up (y), the over or up value is always multiplied by the (√2 ) to find the value of the hypotenuse. For a 30 degree right triangle with  an over (x) of 3 and an up (y) of 4, what is the value of the hypotenuse? Just say.....3, 4, 5! The value of the hypotenuse of a right triangle with sides of 3 and 4 is 5! Prove it!

One thing leads into another. Ke'Ontra and I got busy with the Sine, Cosine and Tangent curves. We got busy with oscillations and the representation of them with sine and cosine waves. We learned that maximum is reached in a sine wave at 90 and 270 degrees (value 1) and minimum values are reached at  0 and 180 degrees. The converse is true with cosine waves. Ke'Ontra took it all in.

I told Ke'Ontra that this was the start of our Electrical training, where tremendous salaries are in the near future!