Law of Sines Day 2- The ambiguous case. Reminder Yesterday we talked in great detail of the 2/3 cases in which you can use law of sines: AAS ASA Today.

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Presentation transcript:

Law of Sines Day 2- The ambiguous case

Reminder Yesterday we talked in great detail of the 2/3 cases in which you can use law of sines: AAS ASA Today we go into more depth on the 3 rd case.

The Ambiguous Case: SSA And it can make one out of you This case includes 2 sides and a non-included angle. Because of this, there can be some issues that must be resolved before solving. The given information (in the order given) limits our knowledge of the supposed triangle. Because of this, there are three possibilities that may occur. To test the triangle for the different possibilities, we build from the ground up.

Method 1: The angle is obtuse (>90 degrees) If the given angle is obtuse, the side opposite of this obtuse angle must be longer than the other given side in order for the triangle to exist. In this case, just use the Law of Sines normally. However, if the given angle is obtuse and the opposite side is not longer than the given side, there is no triangle that can be made using the given information. Thus there is no solution.

Method 2: The angle is acute (< 90 degrees) If the angle is acute, we must test the given information to see if there are any possible triangles to solve for. To do this we build the triangle based on an unknown side that lies on the “ground” In this case, the blue side must reach “the ground” in order for there to be a triangle. If the blue side is not equal to the height, there is no triangle to solve. Unknown side/ “ground” Given Angle Height

To find you height

If the blue side is the same as the height, there only exist one triangle. This is a right triangle and you can use any appropriate method to solve it. If the blue side is longer than the height and the given side, there is only one triangle that you must use law of sines to solve.

If the blue side is longer than the height but shorter than the other side, there are TWO possible triangles. YOU MUST SOLVE FOR BOTH USING THE LAW OF SINES.

Example 1: Given a = 22, b = 12, and A = 42 degrees, find the remaining sides and angles.

Example 2: Given a = 15, b = 25, and A = 85 degrees, find the remaining sides and angles.

Example 3: Given a = 12, b = 31, and A = 20.5 degrees, find the remaining sides and angles.

Homework Worksheet Assignment #2