Math Problem Statement

Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. (If an answer does not exist, enter DNE. Round your answers to one decimal place. Below, enter your answers so that ∠B1 is larger than ∠B2.) a = 33, c = 44, ∠A = 39° ∠B1 = ° ∠B2 = ° ∠C1 = ° ∠C2 = ° b1 =

b2 =

Solution

We are given the following conditions for a triangle:

  • a=33a = 33
  • c=44c = 44
  • A=39\angle A = 39^\circ

We need to solve for the two possible triangles that satisfy these conditions using the Law of Sines.

Step 1: Use the Law of Sines

The Law of Sines states that:

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

From the information provided, we can first calculate the value of asin(A)\frac{a}{\sin(A)}:

33sin(39)330.629352.5\frac{33}{\sin(39^\circ)} \approx \frac{33}{0.6293} \approx 52.5

Now, we can use this to find angle BB (there are two possible cases for BB).

Step 2: Solve for B\angle B

Using the Law of Sines:

33sin(39)=44sin(B)\frac{33}{\sin(39^\circ)} = \frac{44}{\sin(B)}

This simplifies to:

330.6293=44sin(B)sin(B)=44×0.6293330.8492\frac{33}{0.6293} = \frac{44}{\sin(B)} \quad \Rightarrow \quad \sin(B) = \frac{44 \times 0.6293}{33} \approx 0.8492

Now, solving for BB:

B1=sin1(0.8492)58.7B_1 = \sin^{-1}(0.8492) \approx 58.7^\circ

Since the sum of angles in a triangle is 180°, we can calculate CC:

C1=1803958.7=82.3C_1 = 180^\circ - 39^\circ - 58.7^\circ = 82.3^\circ

For the second possible angle B2B_2, we use the fact that sin(B)=sin(180B)\sin(B) = \sin(180^\circ - B), which gives us the second solution for BB:

B2=18058.7=121.3B_2 = 180^\circ - 58.7^\circ = 121.3^\circ

Now, calculating C2C_2:

C2=18039121.3=19.7C_2 = 180^\circ - 39^\circ - 121.3^\circ = 19.7^\circ

Step 3: Solve for sides b1b_1 and b2b_2

Now, using the Law of Sines again to find the lengths of side bb for each case.

For B1=58.7B_1 = 58.7^\circ and C1=82.3C_1 = 82.3^\circ:

b1sin(58.7)=33sin(39)b1=33×sin(58.7)sin(39)33×0.85720.629345.4\frac{b_1}{\sin(58.7^\circ)} = \frac{33}{\sin(39^\circ)} \quad \Rightarrow \quad b_1 = \frac{33 \times \sin(58.7^\circ)}{\sin(39^\circ)} \approx \frac{33 \times 0.8572}{0.6293} \approx 45.4

For B2=121.3B_2 = 121.3^\circ and C2=19.7C_2 = 19.7^\circ:

b2sin(121.3)=33sin(39)b2=33×sin(121.3)sin(39)33×0.83870.629345.0\frac{b_2}{\sin(121.3^\circ)} = \frac{33}{\sin(39^\circ)} \quad \Rightarrow \quad b_2 = \frac{33 \times \sin(121.3^\circ)}{\sin(39^\circ)} \approx \frac{33 \times 0.8387}{0.6293} \approx 45.0

Final Answers

  • B158.7\angle B_1 \approx 58.7^\circ
  • B2121.3\angle B_2 \approx 121.3^\circ
  • C182.3\angle C_1 \approx 82.3^\circ
  • C219.7\angle C_2 \approx 19.7^\circ
  • b145.4b_1 \approx 45.4
  • b245.0b_2 \approx 45.0

Tip:

When applying the Law of Sines, always check if the given conditions can create two possible solutions for the triangle, especially when an angle has two possible corresponding values.

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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)

Theorems

Law of Sines

Suitable Grade Level

Grades 10-12