Today was not overly productive as students had struggled with determining the side lengths and trig ratios for a triangle that had sin(30o) = 1/2. I drew a triangle on the board and labeled one of the angles as the 30o angle. I asked students what possible side lengths could be given the sin(30o) = 1/2. One student volunteered that the opposite side length could be 5 units and the hypotenuse could be 10 units. I wrote these values down.
I then asked what the length of the second leg had to be. Some students jumped at using the Pythagorean theorem while others just sat there. I asked how they could find the missing side length and the students started to realize what they needed to do.
When I asked the class what the value was, one student replied the square root of 75. Others confirmed this. Many students wanted to use their calculator to get an approximate value. I told students this would be a good opportunity to practice simplifying radicals. I had to remind some to use factor trees and we worked through getting a value of 5sqrt(3).
The next struggle came with students calculating the remaining trigonometric ratios. They finally came out of their morning stupor and calculated the values for the cosine and tangent of 30o. I then had to remind them that they needed to calculate all the trigonometric ratios. They then realized they also needed to calculate the sine, cosine, and tangent of 60o. We then checked answers. The results provided more opportunity to simplify radical expressions. We looked at how to re-express 1/sqrt(3), for example.
At this point, I pointed out that they should be able to determine all side lengths and trigonometric ratios from just knowing a single trigonometric ratio. I wrote down that the tan(A) = 3/4 and asked students to determine all the side lengths and trigonometric ratios. This exercise actually went better. There were still a couple of students that were totally confused but the rest of the class appeared to be understanding what was needed and how to proceed.
I then asked students if there were any issues with solving the four equations/proportions. The only problem that students had questions about was the second problem that involved a proportion:
2 / 3 = x / 21. I wrote this on the board and asked students what they attempted or what they could do. One student commented that 3 x 7 = 21. I wrote this expression out for the denominator and noted that we could not just multiply the denominator by 7. Students then saw that the numerator would be 2 x 7 and that x = 14. I also noted they could eliminate the fraction by multiplying by 21 (again noting the multiplicative property of equality). This yielded the result that 21 ( 2 / 3 ) = x or 14 = x.
With this, we were at the end of class. I asked students to complete problems 6-8 on the fourth page of the Relationships with Meaning packet. These are a continuation of the work we had been doing in class. Problem 5 was actually the follow-up problem we had worked on. I told students that it shouldn't take long and that they should not spend more than 30 minutes working through the problems. I am hopeful that tonight's homework will go better than the previous night's assignment.
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