Math Problem Statement

Solve the triangle if possible, given that angle B = 41°, angle C = 28°, and side b = 25.

Solution

To solve this triangle, we can use the Law of Sines since we know:

  1. Angle B=41B = 41^\circ
  2. Angle C=28C = 28^\circ
  3. Side b=25b = 25

We are looking to find the measurements of:

  • Angle AA
  • Sides aa and cc

Step 1: Find Angle AA

Since the sum of the angles in any triangle is 180180^\circ, we can calculate AA as follows:

A=180BC=1804128=111A = 180^\circ - B - C = 180^\circ - 41^\circ - 28^\circ = 111^\circ

Step 2: Use the Law of Sines to Find Side aa

The Law of Sines states:

asin(A)=bsin(B)=csin(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}

Rearrange to solve for aa:

a=bsin(A)sin(B)a = \frac{b \cdot \sin(A)}{\sin(B)}

Substitute the known values:

a=25sin(111)sin(41)a = \frac{25 \cdot \sin(111^\circ)}{\sin(41^\circ)}

Calculating the sine values and solving for aa:

  1. sin(111)0.8910\sin(111^\circ) \approx 0.8910
  2. sin(41)0.6561\sin(41^\circ) \approx 0.6561

a250.89100.656133.9a \approx \frac{25 \cdot 0.8910}{0.6561} \approx 33.9

Step 3: Use the Law of Sines to Find Side cc

Similarly, solve for cc:

c=bsin(C)sin(B)c = \frac{b \cdot \sin(C)}{\sin(B)}

Substitute the known values:

c=25sin(28)sin(41)c = \frac{25 \cdot \sin(28^\circ)}{\sin(41^\circ)}

Calculating the sine values:

  1. sin(28)0.4695\sin(28^\circ) \approx 0.4695
  2. sin(41)0.6561\sin(41^\circ) \approx 0.6561

c250.46950.656117.9c \approx \frac{25 \cdot 0.4695}{0.6561} \approx 17.9

Conclusion

The solution for this triangle is:

  • A111A \approx 111^\circ
  • a34a \approx 34 (rounded to the nearest integer)
  • c18c \approx 18 (rounded to the nearest integer)

Given these calculations, the correct answer choice would be B with:

  • A111A \approx 111^\circ
  • a34a \approx 34
  • c18c \approx 18

Do you have any questions or would like further explanation on any step?


Here are 5 related questions for further practice:

  1. How does the Law of Sines work for finding unknown angles in a triangle?
  2. When is it possible to have two solutions in a triangle using the Law of Sines?
  3. How would you solve for unknown sides if you were given two sides and an angle not between them?
  4. What is the difference between using the Law of Sines and the Law of Cosines?
  5. How can rounding affect the accuracy of triangle solutions?

Tip: When using the Law of Sines, always check if an ambiguous case (SSA) exists, which could lead to two possible solutions.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Triangle Solutions
Law of Sines

Formulas

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

Theorems

Law of Sines

Suitable Grade Level

High School (Grades 10-12)