Notes: Law of Sines Ambiguous Cases LT 1B I can prove the Law of Sines and Law of Cosines, explain when to apply the laws for oblique triangles, and use them to solve problems.

Law of Sines Ambiguous Cases http://youtu.be/pIMsryG0g74

Investigate Given ∆𝐴𝐵𝐶, show that sin 𝐴 𝑎 = sin 𝐵 𝑏 = sin 𝐶 𝑐

The Ambiguous Case (SSA) Situation I: Angle A is obtuse If angle A is obtuse there are TWO possibilities If a ≤ b, then a is too short to reach side c - a triangle with these dimensions is impossible. If a > b, then there is ONE triangle with these dimensions. A B ? a b C = ? c = ? A B ? a b C = ? c = ?

The Ambiguous Case (SSA) Situation II: Angle A is acute First, use SOH-CAH-TOA to find h: A B ? b C = ? c = ? a h Then, compare ‘h’ to sides a and b . . .

The Ambiguous Case (SSA) Situation II: Angle A is acute If a < h, then NO triangle exists with these dimensions. A B ? b C = ? c = ? a h

The Ambiguous Case (SSA) Situation II: Angle A is acute If h < a < b, then TWO triangles exist with these dimensions. A B b C c a h A B b C c a h If we open side ‘a’ to the outside of h, angle B is acute. If we open side ‘a’ to the inside of h, angle B is obtuse.

The Ambiguous Case (SSA) Situation II: Angle A is acute If h < b < a, then ONE triangle exists with these dimensions. Since side a is greater than side b, side a cannot open to the inside of h, it can only open to the outside, so there is only 1 triangle possible! A B b C c a h

The Ambiguous Case (SSA) Situation II: Angle A is acute If h = a, then ONE triangle exists with these dimensions. A B b C c a = h If a = h, then angle B must be a right angle and there is only one possible triangle with these dimensions.

The Ambiguous Case - Summary if angle A is acute find the height, h = b*sinA if angle A is obtuse if a < b  no solution if a > b  one solution if a < h  no solution if h < a < b  2 solutions one with angle B acute, one with angle B obtuse if a > b > h  1 solution If a = h  1 solution angle B is right

The Ambiguous Case (SSA) FIRST SOLUTION: Angle B is acute - this is the solution you get when you use the Law of Sines! A B 15 = b C c a = 12 h 40°

The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 1 SECOND SOLUTION: Angle B is obtuse a = 12 A B 15 = b C c 40° 126.5°

The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 1 SECOND SOLUTION: Angle B is obtuse - use the first solution to find this solution. In the second set of possible dimensions, angle B is obtuse, because side ‘a’ is the same in both solutions, the acute solution for angle B & the obtuse solution for angle B are supplementary. Angle B = 180 - 53.5° = 126.5° a = 12 A B 15 = b C c 40° 1st ‘a’ 1st ‘B’

The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 1 SECOND SOLUTION: Angle B is obtuse Angle B = 126.5° Angle C = 180°- 40°- 126.5° = 13.5° a = 12 A B 15 = b C c 40° 126.5°

The Ambiguous Case (SSA) Situation II: Angle A is acute - EX. 1 (Summary) Angle B = 53.5° Angle C = 86.5° Side c = 18.6 Angle B = 126.5° Angle C = 13.5° Side c = 4.4 a = 12 A B 15 = b C c = 4.4 40° 126.5° 13.5° A B 15 = b C c = 18.6 a = 12 40° 53.5° 86.5°

The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 2 Given a triangle with angle A = 40°, side a = 12 cm and side b = 10 cm, find the other dimensions. A B ? 10 = b C = ? c = ? a = 12 h 40° Since a > b, and h is less than a, we know this triangle has just ONE possible solution - side ‘a’opens to the outside of h.

The Ambiguous Case (SSA) Using the Law of Sines will give us the ONE possible solution: A B 10 = b C c a = 12 40°