# Trig Practice – missing sides #MFM2P #3ActMath

A few days ago I read a blog post by Kate Nowak titled In Defense of Unsexy. She argues that we like to blog about the “sexy”, awesome, new, innovative activities & lessons we’ve tried in our classes. But that there is a lack of sharing the plain old “normal” lessons we still use. I’ve been trying to blog more this semester about the activities we do in MFM2P (grade 10 applied), but am guilty of posting mostly the ones I think are more “sexy”. So I wasn’t going to post what we did the last 2 days as the problems chosen were very basic trig ratio practice problems. But Kate has convinced me that this, too, is worth sharing and reflecting on.

So the problems & the prompts are very basic here. But I adapt them & try to make it engaging by layering the 3 act math model onto them.

WEDNESDAY – MISSING SIDE LENGTH IN A RIGHT TRIANGLE

Act 1: Prompt, notice, wonder, estimate

At first we had a bunch of estimates larger than 8cm. So I asked the students what they thought of the estimates made. A few said that b could not be longer than 8 cm, judging by the diagram which was drawn to scale. So I allowed them to adjust their estimates to what you see above. I still had 2 students that left estimates longer than 8 cm. So does that mean they weren’t listening to our discussion? That they didn’t understand our discussion? That they just didn’t bother to change their estimates? I’m not sure.

Act 2: Solve

All the groups solved this, but I only photographed 1 board (the clearest, nicest looking one) as I wasn’t originally going to blog about this activity.

Act 3:

When the problem is an “unsexy” one like this, I consider act 3 to be the class discussion surrounding the strategies used to solve, the clarity of work on the boards, etc.

Repeat

We then repeated the process for the other missing side.

Act 1:

Act 2:
Half the groups solved using trig, but they often make mistakes when the variable is in the denominator. Another group tried to divide both sides by 8 at that point, and while they got the right final answer, their work did not support it (now I wish I’d photographed each board! lesson learned). This group shown here originally multiplied 8 x 0.9063 to get an answer smaller than 8. They knew their answer was wrong, so I wrote the expression out in green at the top right for them to work with.The other half of the groups used Pythagorean Theorem.

They then started on this Khan Academy homework.

All materials for this lesson are here.

– Laura Wheeler (Teacher @ Ridgemont High School, OCDSB; Ottawa, ON)