I'll go through the objectives one by one and say something about them.
The Law of Sines
The Law of Sines is a very easy way of relating every side in a triangle, along with the angles.
Although sinA/a = sinB/b is equal to their reciprocal, it's better to do
a/sinA = b/sinB = c/sinc, because it also equals 2R, where R is the radius of the circumscribed circle
The proof of the Law of Sines is like so: given a triangle ABC, (counter-clockwise), you can draw an altitude, and label it X. on one side, b sin A = x, and on the other side a sin B = x, so a sin B = b sin A, so a/sinA = b/sin B (it's on khanacademy)
The Law of Cosines
Unlike the Law of Sines, which could yield the incorrect result (sin(x) = sin(180-x)), the Law of Cosines always works. It's kind of similar to the Pythagorean Theorem:
a^2 = b^2 + c^2 - 2bc*cos(A). So, when cos(A) = 0, at 90 degrees, it equals a^2 = b^2 + c^2, a being the hypotenuse.
Proof: Given the same triangle from above, you can draw an altitude going into side c. Now, assume you know angle A (bottom left). The adjacent side (part of C), is bcos(theta). The altitude is bsin(theta). The other part is C - bcos(theta). So the hypotenuse of the right side (a) squared equals the other two sides squared.
a^2 = (bsin(theta))^2 + (c- bcos(theta))^2
a^2 = b^2sin^2(theta) + c^2 - 2bccos(theta) + b^2cos^2theta
a^2 = b^2((sin^2(theta) + cos^2(theta)) + c^2 - 2bc*cos(theta)
a^2 = b^2 + c^2 - 2bc*cos(theta)
Using the Law of Sines and Cosines to solve triangles, including multi-step problems and word problems.
Just do a few problem
Problems that require some basic geometry knowledge (similar triangles, angle bisector theorem, properties of parallelograms, area of a sector of a circle, etc.)
Not much to it, just do it. (rhymes!)
Examples: #2, 4, 7, 8, 9 from the packet called "Using the Law of Sines and Cosines"
Proofs that use the Law of Cosines or Law of Sines
You should be able to prove, for example:
- The Triangle Inequality (use the Law of Cosines)
- The Pythagorean Inequalities (use the Law of Cosines)
(will do later)
- Area of a triangle = 1/2 ab sin(C)
- Heron's Formula - know the proof/derivation
- Using repeated applications of triangle solving to find the area of polygons by splitting them up into triangles: see Foerster, p. 261 #25 and 26
The Ambiguous Case of the Law of Sines
see problem #2 from the quiz
Solving SSS and SAS triangles by being strategic about which angles to solve for first see problem #1 from the quiz and the discussion we had in class on this topic
The formulas for the radii of the circumscribed and inscribed circles of a triangle in terms of the sides and angles of a triangle -- Larson p. 507 #44 (a) and (b)
WILL FINISH COMMENTING ON TOPICS LATER