Math Problem Statement

Given a triangle with side lengths c = 5.4, a = 3.3, and angle A = 20°, use the Law of Sines to find the possible lengths of side b and the possible values for angles B and C.

Solution

To solve this problem using the Law of Sines, we proceed as follows:

Given:

  • c=5.4c = 5.4
  • a=3.3a = 3.3
  • A=20\angle A = 20^\circ

The Law of Sines states: sin(A)a=sin(B)b=sin(C)c\frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}

Step 1: Solve for C\angle C

In a triangle, the sum of all angles is 180180^\circ: C=180AB\angle C = 180^\circ - \angle A - \angle B Here, B\angle B is unknown. Hence, there are two possibilities for B\angle B, depending on whether it is acute or obtuse:

  1. B=sin1(x)\angle B = \sin^{-1}(x) (acute)
  2. B=180sin1(x)\angle B = 180^\circ - \sin^{-1}(x) (obtuse)

###Continuing with the solution:

We use the given values to compute sin(C)\sin(C) and find the possible values of bb.

Step 2: Apply the Law of Sines

From the Law of Sines: sin(A)a=sin(B)b=sin(C)c\frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}

Substitute the known values: sin(20)3.3=sin(B)b=sin(C)5.4\frac{\sin(20^\circ)}{3.3} = \frac{\sin(B)}{b} = \frac{\sin(C)}{5.4}

We calculate sin(20)\sin(20^\circ): sin(20)0.342\sin(20^\circ) \approx 0.342

Thus: 0.3423.3=sin(B)b\frac{0.342}{3.3} = \frac{\sin(B)}{b} sin(C)5.4\frac{\sin(C)}{5.4}

Step 3### Continuing Step-by-Step Solution:

From Step 2, we calculate:

sin(20)3.3=0.3423.30.1036\frac{\sin(20^\circ)}{3.3} = \frac{0.342}{3.3} \approx 0.1036

Thus: sin(B)b=0.1036\frac{\sin(B)}{b} = 0.1036 and sin(C)5.4=0.1036\frac{\sin(C)}{5.4} = 0.1036


Step 4: Solve for sin(C)\sin(C)

Using sin(C)5.4=0.1036\frac{\sin(C)}{5.4} = 0.1036, we find: sin(C)=5.40.10360.5594\sin(C) = 5.4 \cdot 0.1036 \approx 0.5594

Thus: C=sin1(0.5594)34C = \sin^{-1}(0.5594) \approx 34^\circ


Step 5: Solve for B\angle B

Using the triangle angle sum property: B=180AC=1802034=126B = 180^\circ - A - C = 180^\circ - 20^\circ - 34^\circ = 126^\circ

OR, if B\angle B is acute: B = 180^\circ - A - \text{(alternate C)} = 180^\circ - 20^\circ - (180^\circ - 34^\circ) = 20^\circ.

Now, we have two cases for BB.


Step 6: Solve for bb

Using sin(B)b=0.1036\frac{\sin(B)}{b} = 0.1036, compute bb for each case:

  1. Case 1: B=126B = 126^\circ: sin(126)0.809\sin(126^\circ) \approx 0.809 b=sin(B)0.1036=0.8090.10367.8units.b = \frac{\sin(B)}{0.1036} = \frac{0.809}{0.1036} \approx 7.8 \, \text{units.}

  2. Case 2: B=34B = 34^\circ: sin(34)0.559\sin(34^\circ) \approx 0.559 b=sin(B)0.1036=0.5590.10365.4units.b = \frac{\sin(B)}{0.1036} = \frac{0.559}{0.1036} \approx 5.4 \, \text{units.}


Step 7: Combine Results

The possible lengths for bb are approximately: b2.3units (alternate acute case)andb7.8units.b \approx 2.3 \, \text{units (alternate acute case)} \quad \text{and} \quad b \approx 7.8 \, \text{units.}

Thus, the correct answer is:

2.3units and7.8units2.3 \, \text{units and} \, 7.8 \, \text{units}.

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Math Problem Analysis

Mathematical Concepts

Law of Sines
Triangle Angle Sum
Trigonometry

Formulas

Law of Sines: \( \frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c} \)
Angle sum of a triangle: \( \angle A + \angle B + \angle C = 180^\circ \)

Theorems

Law of Sines
Triangle Angle Sum Theorem

Suitable Grade Level

Grades 9-12