Presentation on theme: "Law of Sines The Ambiguous Case Section 6-1. In geometry, triangles can be uniquely defined when particular combinations of sides and angles are specified."— Presentation transcript:
Law of Sines The Ambiguous Case Section 6-1
In geometry, triangles can be uniquely defined when particular combinations of sides and angles are specified … this means we can solve them! Angle Side Angle - ASA Angle Angle Side – AAS We solved these using Law of Sines Then there are these theorems … Side Side Side – SSS Side Angle Side – SAS We’ll soon solve these using the Law of Cosines (section 6.2)
There’s one left … Side Side Angle - SSA There’s a problem with solving triangles given SSA … You could find… 1.No solution 2.One solution 3.Two solutions In other words … its AMBIGUOUS … unclear Let’s take a look at each of these possibilities.
Remember now … the information we’re given is two consecutive sides and the next angle … If this side isn’t long enough, then we can’t create a triangle … no solution So, then, what is the “right” length so we can make a triangle? An altitude … 90 degree angle … a RIGHT triangle! Turns out this is an important calculation … it’s a = b sin θ If a = b sin θ, then there is only one solution for this triangle. The missing angle is the complement to θ The missing side can be found using Pythagorean theorem of trigonometry. θ a b
What if side a is a little too long … what would that look like? a > b sin θ This leg can then either swing left … or right. So? … which one of these triangles do you solve? … BOTH! First, solve the acute triangle … and find angle B by Law of Sines! Then solve for the remaining parts of the acute triangle θ a b B
Lastly, solve the obtuse triangle … This next step is critical … angle B’ is ALWAYS the supplement to angle B. B’ = 180 – m< B Next, solve the remaining parts of the obtuse triangle. θ b a BB’
Here’s the last scenario while θ is acute … What if side a is larger than side b? In this case, only one triangle can exist … an acute triangle which can easily be solved using law of sines. Too long to create a triangle on this side. θ a