Math Problem Statement

Two sides and an angle are given below. Determine whether the given information results in one​ triangle, two​ triangles, or no triangle at all. Solve any resulting​ triangle(s). b equals 8 commab=8, c equals 9 commac=9, Upper B equals 30 degreesB=30° Question content area bottom Part 1 Select the correct choice below​ and, if​ necessary, fill in the answer boxes to complete your choice. ​(Type an integer or decimal rounded to two decimal places as​ needed.) A. A single triangle is​ produced, where Calmost equals≈enter your response heredegrees°​, Aalmost equals≈enter your response heredegrees°​, and aalmost equals≈enter your response here. B. Two triangles are​ produced, where the triangle with the smaller angle C has Upper C 1C1almost equals≈enter your response heredegrees°​, Upper A 1A1almost equals≈enter your response heredegrees°​, and a 1a1almost equals≈enter your response here​, and the triangle with the larger angle C has Upper C 2C2almost equals≈enter your response heredegrees°​, Upper A 2A2almost equals≈enter your response heredegrees°​, and a 2a2almost equals≈enter your response here. C. No triangles are produced.

Solution

To determine whether the given information results in one triangle, two triangles, or no triangle at all, we need to apply the Law of Sines to solve for the unknown angles and sides. Given are:

  • b=8b = 8
  • c=9c = 9
  • B=30B = 30^\circ

Step 1: Use the Law of Sines to solve for angle CC

bsin(B)=csin(C)\frac{b}{\sin(B)} = \frac{c}{\sin(C)} Substitute the known values: 8sin(30)=9sin(C)\frac{8}{\sin(30^\circ)} = \frac{9}{\sin(C)} Since sin(30)=0.5\sin(30^\circ) = 0.5, this simplifies to: 80.5=9sin(C)\frac{8}{0.5} = \frac{9}{\sin(C)} 16=9sin(C)16 = \frac{9}{\sin(C)} Now solve for sin(C)\sin(C): sin(C)=916=0.5625\sin(C) = \frac{9}{16} = 0.5625

Step 2: Check for the number of possible triangles

Since sin(C)=0.5625\sin(C) = 0.5625, this gives us two possible solutions for angle CC:

  • C1=sin1(0.5625)34.22C_1 = \sin^{-1}(0.5625) \approx 34.22^\circ
  • C2=18034.22145.78C_2 = 180^\circ - 34.22^\circ \approx 145.78^\circ

Case 1: C1=34.22C_1 = 34.22^\circ

Now, to find angle AA, use the fact that the sum of the angles in a triangle is 180°: A1=180BC1=1803034.22=115.78A_1 = 180^\circ - B - C_1 = 180^\circ - 30^\circ - 34.22^\circ = 115.78^\circ This is a valid triangle since all angles are positive and less than 180°.

Case 2: C2=145.78C_2 = 145.78^\circ

For the second triangle: A2=180BC2=18030145.78=4.22A_2 = 180^\circ - B - C_2 = 180^\circ - 30^\circ - 145.78^\circ = 4.22^\circ This is also a valid triangle, though A2A_2 is quite small.

Step 3: Solve for side aa

Using the Law of Sines again to solve for side aa in both cases:

For the first triangle: a1sin(A1)=bsin(B)\frac{a_1}{\sin(A_1)} = \frac{b}{\sin(B)} a1sin(115.78)=8sin(30)\frac{a_1}{\sin(115.78^\circ)} = \frac{8}{\sin(30^\circ)} a10.9063=80.5\frac{a_1}{0.9063} = \frac{8}{0.5} a1=8×0.90630.514.50a_1 = \frac{8 \times 0.9063}{0.5} \approx 14.50

For the second triangle: a2sin(A2)=bsin(B)\frac{a_2}{\sin(A_2)} = \frac{b}{\sin(B)} a2sin(4.22)=80.5\frac{a_2}{\sin(4.22^\circ)} = \frac{8}{0.5} a20.0736=16\frac{a_2}{0.0736} = 16 a2=16×0.07361.18a_2 = 16 \times 0.0736 \approx 1.18

Conclusion:

There are two triangles formed:

  • For the first triangle:

    • C134.22C_1 \approx 34.22^\circ
    • A1115.78A_1 \approx 115.78^\circ
    • a114.50a_1 \approx 14.50
  • For the second triangle:

    • C2145.78C_2 \approx 145.78^\circ
    • A24.22A_2 \approx 4.22^\circ
    • a21.18a_2 \approx 1.18

Would you like further explanation on any part of the process?


Related Questions:

  1. What is the Law of Sines, and how is it applied to non-right triangles?
  2. Can there be situations where no triangle is formed when given two sides and an angle?
  3. How do you handle ambiguous cases in trigonometry?
  4. What is the difference between SSA, SAS, and ASA triangle configurations?
  5. How do you find the area of a triangle when given two sides and an angle?

Tip: The Law of Sines is particularly useful for solving triangles when two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA) are known.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Law of Sines
Triangle Properties

Formulas

Law of Sines: (a/sin(A)) = (b/sin(B)) = (c/sin(C))
Sum of angles in a triangle: A + B + C = 180°

Theorems

Law of Sines
Angle-Side-Angle (ASA) Theorem
Ambiguous Case (SSA) Theorem

Suitable Grade Level

Grades 10-12