More trigonometry

The unit circle, compound angles etc took longer than planned on Friday. As I said in class, these topics come up infrequently, so if you are aiming to pass this question, make sure that you are able to use sin=opp/hyp etc, sine rule, cosine rule and area of triangle.

I forgot to post the 1-page overview of trigonometry last time.



Also, I will mention this in tomorrow's class, but there is one thing you should be aware of when using the Sine rule:
If you use your calculator to work out the inverse sine of 0.9897, your calculator will tell you it is 81.7 degrees ... which it is. But, it is also the sine of 97.3 degrees. You know from the discussion of unit circle that there are two possible angles that have the same sine, cosine and tan. Our calculators always return the angle that is between 0 and 90 degrees only, not the obtuse alternative.
Look at 2003 P2 Q5 for an example where this can cause problems.
If you solve (c) (ii) by putting
Sin60/7 = SinA/8
you will get the wrong answer.
The problem here is that the shorter side of the triangle (3) is not used in this equation and so the triangle where we have a known angle (60) and two other sides (7 and 8) is not a well defined triangle. In effect, it is ASS (angle, side, side) rather than SAS (side, angle, side)! (remember the 4 conditions for congruence of triangles that you learned: SSS,SAS,ASA,RHS)
You have to use the acute angle to work it out.
Sin60/7 = SinB/3 ... which gives you 22 degrees for that angle and leaves 98 degrees for the obtuse angle (60 + 22 + 98 = 180).

How do you know which angle to go for? A triangle can have at most 1 obtuse angle (> 90 degrees) and that will be opposite the longest side.
If you have a choice, solve the acute angle, opposite the shortest side, first.