College

Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. There may be two triangles, one triangle, or no triangles that satisfy the given conditions. (If an answer does not exist, enter DNE. Enter all angles in degrees. Round your answers to one decimal place. Below, enter your answers so that [tex]B_1[/tex] is larger than [tex]B_2[/tex].)

Given:
- [tex]a = 36[/tex]
- [tex]c = 46[/tex]
- [tex]A = 35^\circ[/tex]

Find:
- [tex]B_1[/tex]
- [tex]C_1[/tex]
- [tex]C_2[/tex]
- [tex]b_1[/tex]
- [tex]b_2[/tex]

Answers:
- [tex]B_1 = 101.2^\circ[/tex]
- [tex]B_2 = \text{x}^\circ[/tex] (DNE if no solution)
- [tex]C_1 = 47.4^\circ[/tex]
- [tex]C_2 = 132.6^\circ[/tex]
- [tex]b_1 = 97.6[/tex]
- [tex]b_2 = 13.2[/tex] (DNE if no solution)

Answer :

Answer:

Here's how to solve this triangle problem using the Law of Sines:

**1. Find angle C₁:**

* Law of Sines: sin(C)/c = sin(A)/a

* sin(C₁)/46 = sin(35°)/36

* sin(C₁) = (46 * sin(35°))/36

* sin(C₁) ≈ 0.735

* C₁ = arcsin(0.735) ≈ 47.4°

**2. Find angle B₁:**

* The angles in a triangle add up to 180°.

* B₁ = 180° - A - C₁

* B₁ = 180° - 35° - 47.4°

* B₁ ≈ 97.6°

**3. Find side b₁:**

* Law of Sines: b/sin(B) = a/sin(A)

* b₁/sin(97.6°) = 36/sin(35°)

* b₁ = (36 * sin(97.6°))/sin(35°)

* b₁ ≈ 61.8

**4. Check for a second triangle (ambiguous case):**

* Since sin(C) can be positive in both the first and second quadrants, there's a possibility of a second angle C₂.

* C₂ = 180° - C₁

* C₂ = 180° - 47.4°

* C₂ ≈ 132.6°

**5. Check if the second triangle is possible:**

* A + C₂ = 35° + 132.6° = 167.6°

* Since 167.6° < 180°, a second triangle *is* possible.

**6. Find angle B₂:**

* B₂ = 180° - A - C₂

* B₂ = 180° - 35° - 132.6°

* B₂ ≈ 12.4°

**7. Find side b₂:**

* Law of Sines: b/sin(B) = a/sin(A)

* b₂/sin(12.4°) = 36/sin(35°)

* b₂ = (36 * sin(12.4°))/sin(35°)

* b₂ ≈ 13.2

**Answers (B₁ > B₂):**

* B₁ ≈ 97.6°

* C₁ ≈ 47.4°

* b₁ ≈ 61.8

* B₂ ≈ 12.4°

* C₂ ≈ 132.6°

* b₂ ≈ 13.2

Step-by-step explanation:

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