Investigstion

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 ) a=9 C A y=33.748 B=34o b Solution#1 y≈34o A=30o And as I stated previously, in order to find the second possible solution we will have to replace θ1 by 34 in the equation 180 - θ1= θ2. b=10 C 180 - 34= θ2 b a a=9 180 - 34= 146 B c B=146o 146 = θ2 So to make sure that 146 is a solution we will have to add angle B and angle A and see if the answer is >180. 146+30=176. Therefore there is two solutions for when x=9. So we can conclude that this is an ambiguous case. After comparing the fist two triangles , we can state that whenever side ‘’A’’ is longer than side ‘’b’’, they will be only one solution possible.But if side ‘’b’’ is longer then side ‘’a’’ there will be two solution possible. Example#2 (where x=11) y=sin-1( 5x ) y=sin-1( 511 ) y=27.035 Solution#1 y≈27o 180-27=153 Making sure 153 is a solution 153o+30=183 Therefore there is one solution. Here are additional examples that help support the statement previously made. Example#1 (where x = 6) y=sin-1( 5x ) y=sin-1 56 y=56 Solution#1 y≈56o 180 - 56= 124 Solution#2 y=124 So we can see that my statement has been proven. Because when I replaced ‘’x’’ by 6 which was smaller then length ‘’b’’ there was two solutions possible. But when I replaced ‘’x’’ by 11 which is greater then the length ‘’b’’ there was only one solution possible. Let’s solve for ’’ y’’ when x=4 and see how many possible solutions. B= no answer a A y=sin-1 5x y=sin-1 54 b y=sin-1 (1.25) y=sin-1 (1.25) There is no answer. This triangle doesn’t exist. As we can see, it is clear that when x=4 there is no triangle possible. We can see there is no solution in two ways. First we can see that we cannot find a solution algebraically. As we all know, the maximum sine value is sin90 which gives us 1 and the inverse of 1 is 90o. The maximum inverse value is 1. As we can see in the question above we had to find the inverse value of 54 . So far everything seems to be good. But we quickly realize that there is a problem when you divide them because you get 1,25. And as I stated before, the maximum inverse value is 1. So according to the statement made, there is no triangle possible for x=4. This statement tells us that there is another restriction for the value of ‘’x’’ in order for the triangle to be an ambiguous case. The first one that we found was that ‘’x’’ couldn’t be greater than 10. And now the second one was that ‘’x’’ couldn’t be less than 5. The second statement was made because 55 gives us 1 which is the maximum inverse value. The following examples are going to give us the exact restriction for ‘’x’’ in this triangle. Let’s solve for ’’ y’’ when x=10 and find the restriction. y=sin-1 5x y=sin-1 510 y=sin-1 0.5 y=30 180 - θ1= θ2 180 - 30= 150 Is 150+30 less than 180' No, so when x=10 there one solution. Let’s solve for ’’ y’’ when x=5 and find the restriction. y=sin-1 5x y=sin-1 55 y=sin-1 1 y=90 From this example we can conclude that if we want The triangle to have an ambiguous case 5