Math Problem Statement

the ambiguous case of sine law

Solution

The "ambiguous case" of the Sine Law occurs when solving for a triangle using two sides and a non-included angle (SSA configuration). This situation can lead to three different outcomes:

  1. No Triangle: Sometimes, no valid triangle can be formed.
  2. One Triangle: There is exactly one possible triangle.
  3. Two Triangles: There are two distinct possible triangles.

Understanding the Ambiguous Case:

Suppose you are given:

  • Side aa
  • Side bb
  • Angle AA (opposite to side aa)

You want to determine angle BB using the Sine Law: asinA=bsinB\frac{a}{\sin A} = \frac{b}{\sin B} This leads to: sinB=bsinAa\sin B = \frac{b \cdot \sin A}{a}

Outcomes Based on the Value of sinB\sin B:

  1. No Triangle (Impossible Case):

    • If bsinAa>1\frac{b \cdot \sin A}{a} > 1, then sinB>1\sin B > 1, which is impossible since the sine of an angle cannot exceed 1. Therefore, no triangle can be formed.
  2. One Triangle:

    • If bsinAa=1\frac{b \cdot \sin A}{a} = 1, then sinB=1\sin B = 1, which means B=90B = 90^\circ. In this case, only one right triangle can be formed.

    • Alternatively, if 0<bsinAa<10 < \frac{b \cdot \sin A}{a} < 1 and side aa is longer than side bb, only one triangle can be formed with B=sin1(bsinAa)B = \sin^{-1} \left(\frac{b \cdot \sin A}{a}\right).

  3. Two Triangles:

    • If 0<bsinAa<10 < \frac{b \cdot \sin A}{a} < 1 and side aa is shorter than side bb, two different angles BB can satisfy the equation:
      • B1=sin1(bsinAa)B_1 = \sin^{-1} \left(\frac{b \cdot \sin A}{a}\right)
      • B2=180B1B_2 = 180^\circ - B_1

    In this case, two distinct triangles can be formed, because both B1B_1 and B2B_2 are valid angles that add up with AA to form a triangle.

Example Problem:

Given:

  • A=30A = 30^\circ
  • a=7a = 7 units
  • b=10b = 10 units

Find angle BB using the Sine Law: sinB=bsinAa=10sin307=100.57=570.7143\sin B = \frac{b \cdot \sin A}{a} = \frac{10 \cdot \sin 30^\circ}{7} = \frac{10 \cdot 0.5}{7} = \frac{5}{7} \approx 0.7143 Now, B1=sin1(0.7143)45.6B_1 = \sin^{-1}(0.7143) \approx 45.6^\circ.

Since 0<sinB<10 < \sin B < 1, there is also a second solution: B2=180B118045.6=134.4B_2 = 180^\circ - B_1 \approx 180^\circ - 45.6^\circ = 134.4^\circ

So, two triangles are possible:

  • One with B145.6B_1 \approx 45.6^\circ
  • One with B2134.4B_2 \approx 134.4^\circ

Do you have any questions or need further clarification?

Here are 5 related questions to deepen your understanding:

  1. How does the ambiguous case affect the overall solution to a triangle problem?
  2. What are some real-life scenarios where the ambiguous case might occur?
  3. How do you determine the length of the third side in both possible triangles?
  4. Can the ambiguous case occur in right triangles? Why or why not?
  5. How would you handle the ambiguous case if angle BB were obtuse in the first solution?

Tip: When dealing with the ambiguous case, always sketch the potential triangles to visualize the situation better. This can help in understanding whether one or two triangles are possible.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Triangle Geometry

Formulas

Sine Law

Theorems

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Suitable Grade Level

High School