In the following investigation, we will learn about the ambiguous case in trigonometry. When you are solving a triangle, where you know two sides on which is adjacent to the known angle and the other that is opposite to the known angle is called the ambiguous case. In this investigation we will learn when the ambiguous case has zero, one or two different solutions.

In the first case of this investigation (case a), we will be working mostly with acute triangles. In the second case of this investigation (case b), we will be working mostly with obtuse triangles.

Case A

B

For the following triangle abc where A=30o, b=10 and a=x we will be replacing ‘’x’’ and ‘’b’’ by multiple values in order to see whether the triangle has 1 or 2 possible solutions. But before doing this we will need to find the equations that establishes the relationship between ‘’x’’ and ‘’y’’. In order to find the equation we will have to use the sine law as shown:

A=30o

sinAa=sinBb

b=10

a

sin30x=siny10

B=y

C

A

a=x

10sin30x=siny1

b

10sin30x=siny

y=sin-1 ( 10sin30x )

y=sin-1( 5x )

The relation between ‘’x and ‘’y’’ are both shown graphically and algebraically.

y=sin-1( 5x )

B

Let’s solve for ’’ y’’ when x=14 and see if there is 1 or 2 possible solutions.

c

A=30o

y=sin-1( 5x )

a

b=10

y=sin-1( 514 )

a=14

C

A

y=20.92483

B=21o

b

Solution#1 y≈21o

In order to find the second solution possible we will have to replace θ1 by the angle that we found such as 21 in the equation 180 - θ1= θ2.

180-21= θ2

180-21=159

B

So if there is a second solution the solution would 159o.

A=30o

c

a

b=10

a=14

C

A

B=159o ?

b

In a triangle all three angles have to add up to 180. But when you add up angle B and angle A it’s already gives you 189. So B is definitely not 159. Therefore there is only one solution for this triangle where x=14.

Let’s solve for ’’ y’’ when x=9 and see how many possible solutions.

B

c

A=30o

y=sin-1( 5x )

b=10

a

y=sin-1( 59...

In the first case of this investigation (case a), we will be working mostly with acute triangles. In the second case of this investigation (case b), we will be working mostly with obtuse triangles.

Case A

B

For the following triangle abc where A=30o, b=10 and a=x we will be replacing ‘’x’’ and ‘’b’’ by multiple values in order to see whether the triangle has 1 or 2 possible solutions. But before doing this we will need to find the equations that establishes the relationship between ‘’x’’ and ‘’y’’. In order to find the equation we will have to use the sine law as shown:

A=30o

sinAa=sinBb

b=10

a

sin30x=siny10

B=y

C

A

a=x

10sin30x=siny1

b

10sin30x=siny

y=sin-1 ( 10sin30x )

y=sin-1( 5x )

The relation between ‘’x and ‘’y’’ are both shown graphically and algebraically.

y=sin-1( 5x )

B

Let’s solve for ’’ y’’ when x=14 and see if there is 1 or 2 possible solutions.

c

A=30o

y=sin-1( 5x )

a

b=10

y=sin-1( 514 )

a=14

C

A

y=20.92483

B=21o

b

Solution#1 y≈21o

In order to find the second solution possible we will have to replace θ1 by the angle that we found such as 21 in the equation 180 - θ1= θ2.

180-21= θ2

180-21=159

B

So if there is a second solution the solution would 159o.

A=30o

c

a

b=10

a=14

C

A

B=159o ?

b

In a triangle all three angles have to add up to 180. But when you add up angle B and angle A it’s already gives you 189. So B is definitely not 159. Therefore there is only one solution for this triangle where x=14.

Let’s solve for ’’ y’’ when x=9 and see how many possible solutions.

B

c

A=30o

y=sin-1( 5x )

b=10

a

y=sin-1( 59...