5.5 The Ambiguous Case

Although the sine law might be easier to calculate than the cosine law, sometimes the sine law lies to you.

Ex 1 (a) Find the length of B

A = 30ֹ°, b = 9, a = 5

What happens here is that there are two possible triangles that can appear with these dimensions. When you know two sides and a non-contained acute angle, there are three possibilities:

0 solutions – impossible triangle

a < bSinA

1 solution – right triangle

a = bSinA

2 solutions – acute/obtuse triangle (The Ambiguous Case)

a > bSinA

Prove that this is an ambiguous triangle.

Ex 1 (b) Solve the above triangle and get the acute solution

Before we get the obtuse solution, take a look at this:

sin30° sin 150° sin 80° sin100° sin10° = sin?

(c) Solve the triangle and get the obtuse solution.

The easy way to spot an ambiguous triangle: (in order of easiness to spot)

1) You need to be given two sides and a non-contained acute angle. (otherwise known as the acute ASS rule)

2) The side opposite the given angle needs to be the shorter of the two given sides.

3) a > bSinA as given earlier

Ex 2 Spot the ambiguous triangle by the given information.

(a) A = 54°, B = 15°, a = 12

(b) A = 71°, b = 5, c = 7

(c) A = 40°, b = 10, a = 7

(d) A = 100°, b = 7, a = 5

When asked to solve an ambiguous triangle, you must solve for both answers. In a word problem situation, you can solve for the relevant solution only.

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