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9.3 The Law of Sines9.3 The Law of SinesAMBIGUOUS CASEAMBIGUOUS CASE
Then tell whether the arc crosses the other ray of A and, if so, in how many points.
a) Compass at C and opened to 4 cm
b) Compass at C and opened to 5 cm
c) Compass at C and opened to 6 cm
Activity 1 shows that when you are given the lengths of two sides of a triangle and the measure of a nonincluded angle (SSA), it may be possible to construct no triangle, one triangle, or two triangles. For this reason the SSA situation is called the ambiguous case.
Activity 1For this activity, use a ruler, compass, and protractor.
Draw A with measure 30o.
Along one ray of A, locate point C 10 cm from point A.
For each of the following compass settings, draw a large arc.
no
yes, 1
yes, 2
Activity 2
Show that your answers to Activity 1 agree with what the law of sines would give in each of the following SSA situations.
No solution
90
56.4 , 123.6
a) If A = 30o, b = 10, and a = 4, find B.
b) If A = 30o, b = 10, and a = 5, find B.
c) If A = 30o, b = 10, and a = 6, find B.
Ambiguous Case for Acute Angle A
Ambiguous Case for Obtuse Angle A
Ambiguous Case
• If given the lengths of two sides and the angle opposite one of them, it is possible that 0, 1, or 2 such triangles exist.
• Some basic facts that should be kept in mind:
1. For any angle , –1 sin 1, if sin = 1, then = 90o and the triangle is a right triangle.
2. sin = sin(180o – ).
3. The smallest angle is opposite the shortest side, the largest angle is opposite the longest side, and the middle-value angle is opposite the intermediate side (assuming unequal sides).
Possible Outcomes
Outcome 1: If A is acute and a < b.
A
C
B
ba
c
h = b sin A
a) If a < b sinA
A
C
B
b
a
c
h
NO SOLUTION
In this case, when applying the Law of Sines, you may get sinB > 1.
Outcome 1: If A is acute and a < b
A
C
B
b a
c
h = b sin A
b) If a = b sinA
A
C
B
b= a
c
h
1 SOLUTION
Possible Outcomes
A
C
B
b a
c
h = b sin A
c) If a > b sinA
A
C
B
b
c
h
2 SOLUTIONS
a a
B
180 -
Outcome 1: If A is acute and a < b.
Possible Outcomes
Outcome 2: If A is obtuse and a > b
C
A B
a
b
c
ONE SOLUTION
Possible Outcomes
Outcome 2: If A is obtuse and a ≤ b
C
A B
a
b
c
NO SOLUTION
Possible Outcomes
In this case, when applying the Law of Sines, you may get sinB > 1.
Solving the Ambiguous Case: No Such Triangle
Example 3: Solve the triangle ABC if B = 55°40´, b = 8.94 meters, and a = 25.1 meters.
[Solution] Use the law of sines to find A.
Since sin A cannot be greater than 1, the triangle does not exist.
3184379.2sin94.8
'4055sin1.25
sin
sinsin
A
A
bB
aA
Case 2 Two sides and one angle not included between the sides known:
»SSA
Example 4:
Solve the triangle ABC if A = 55.3o, a = 22.8 feet, and b = 24.9 feet.
[Solution]
Solving the Ambiguous Case: Two Triangles
1.1169.63180
9.638978678.sin
9.24sin
8.223.55sin
sinsin
2
1
B
BB
B
bB
aA
Case 2 Two sides and one angle not included between the sides known:
»SSA
To see if B2 = 116.1o is a valid possibility, add 116.1o to the measure of A: 116.1o + 55.3o = 171.4o. Since this sum is less than 180o, it is a valid triangle.
Now separate the triangles into two: AB1C1 and AB2C2. 8.609.633.55180180 11 BAC
feet 2.248.60sin3.55sin
8.22
sinsin
1
1
1
1
c
c
Cc
Aa
Solving the Ambiguous Case: No Such Triangle
60.8o
Now solve for triangle AB2C2.
6.81.1163.55180180 22 BAC
feet 15.46.8sin3.55sin
8.22
sinsin
2
2
2
2
c
c
Cc
Aa
Solving the Ambiguous Case: No Such Triangle
Number of Triangles Satisfying the Ambiguous Case
Let sides a and b and angle A be given in triangle ABC. (The law of sines can be used to calculate sin B.)
1. If sin B > 1, then no triangle satisfies the given conditions.
2. If sin B = 1, then one triangle satisfies the given conditions and B = 90°.
3. If 0 < sin B < 1, then either one or two triangles satisfy the given conditions
(a) If sin B = k, then let B1 = sin-1 k and use B1 for B in the first triangle.
(b) Let B2 = 180° – B1. If A + B2 < 180°, then a second triangle exists. In this case, use B2 for B in the second triangle.
Assignment
P. 348 #7, 13, 17, 21, 22
