Hints for The Ambiguous Case worksheet
1.
The testing set-up would be:
. When you calculate 34 x sin 50, you get 26. That
makes the first half of the test say: 26 < 9. Obviously, that is not true. Therefore, the given value of
9, that’s supposed to be across from the angle 50, is too short and you cannot make a triangle.
Answer: no triangle is possible.
2. The testing set-up would be:
. Obviously, the second half of that statement is
not true. It’s really that 36 > 17. Therefore, the given 36 that’s across from the angle 96 is long
enough to intersect the unknown third side exactly once. Answer: there is 1 triangle.
3. The testing setup would be:
. When you calculate 29 x sin 129, you get 22.5,
so now the first half of the test says 22.5 < 7. Since that is not true, 7 is too short, and we cannot
reach across the angle to make a triangle. Answer: no triangle.
4. The test would say:
test now says:
. When you calculate 25 x sin 50, you get 19.15, so the
. Since this is true, there are two possible triangles.
5. The test would say:
. The second half is obviously not true, since 33 > 18.
Therefore, the 33 is long enough to intersect the unknown third side exactly once and give one
triangle.
6. The test would say:
. When you calculate 22 x sin 28, you get 10.33, which
makes the statement true. Therefore, there are two triangles possible.
7. The test would say:
. This is a special case --- the second half fails, since 34 =
34, not 34 < 34. If the angle had been acute, then I could make 1 triangle, like I could on #2 or #5.
But the angle is obtuse, which means that the picture would look something like this:
34
34
136
Now, that might look okay, but consider --- that’s an isosceles
triangle. The angles across from both of the 34’s should be
equal. But there’s no way both angles could be 136. So this
picture is not possible. There are no triangles to match these
numbers.
8. The test would say:
. Calculating 33 x sin 28 gives 15.5, so the statement is
true and there are two possible triangles.
FOLLOW-UP: SOLVE THE TRIANGLES for questions 2, 4, 6 and 8. (Remember, “solve the triangle” means
“find the missing bits”.)
#2, part b) set up a Law of Sines proportion involving sin 96 over 36, and sin B over 17. Calculate sin B,
use an inverse to find B. That is the only possible value of B, since only 1 triangle is possible. Calculate
angle C by subtracting 180 – 96 – what you got for B. Set up a Law of Sines proportion involving sin 96
over 35 on one side, and sin whatever-you-got-for-angle-C, over little c, on the other side. Solve for
little c.
The other problems follow a similar pattern, although they switch around the letters. But there will be
two possible triangles, so you will have two possible values for the missing angles. Organize your work
into two columns, like this:
Solve this proportion for the missing angle
B1 = answer you found from your first proportion
B2 = the supplement, 180 – B1
C1 = 180 – 96 – whatever B1 is
C2 = 180 – 96 – whatever B2 is
Set up a Law of Sines proportion:
Set up a Law of Sines proportion:
Solve for little c
Solve for little c
The letters will keep changing, but bear in mind that all three proportions have the same first half, and
it’s the pair of numbers from the original problem that share a letter. (They will be the only angle-side
letter pair given.)
#4, part b) set up a proportion involving sin 50 over 24, and sin A over 25. Solve for angle A. Label it A1,
then go find the supplement ‘cause that’s A2. Find the two possible versions of C (by subtracting 180 –
given angle of 50 – both possible values of A). Set up proportions involving sin 50 over 24 and the
different options for angle C. Solve both proportions for little c.
#6, part b) same idea as #4, but with sin 28 instead of sin 50, 19 instead of 24, and 22 instead of 25.
#8, part b) same idea as #4, but with sin 28 instead of sin 50; 19 instead of 24, and 33 instead of 25.