Math Problem Statement

At least one of your answers is incorrect. To solve the​ triangle, use the Law of Sines. For a triangle with sides​ a, b, c and opposite angles​ A, B,​ C, respectively,

StartFraction sine Upper A Over a EndFraction equals StartFraction sine Upper B Over b EndFraction equals StartFraction sine Upper C Over c EndFractionsinAa=sinBb=sinCc.

​Also, note that

Aplus+Bplus+Cequals=180degrees°.

Solution

The Law of Sines is a useful tool for solving a triangle when you are given either:

  1. Two angles and one side (AAS or ASA case)
  2. Two sides and a non-included angle (SSA case)

The Law of Sines states:

sinAa=sinBb=sinCc\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}

where:

  • AA, BB, and CC are the angles of the triangle.
  • aa, bb, and cc are the sides opposite these angles, respectively.

Additionally, the sum of the angles in any triangle is:

A+B+C=180A + B + C = 180^\circ

How to apply the Law of Sines:

  1. If you know two angles and one side:

    • Find the third angle using the equation A+B+C=180A + B + C = 180^\circ.
    • Then use the Law of Sines to find the unknown sides.
  2. If you know two sides and one angle (SSA case):

    • Use the Law of Sines to find the other angle.
    • Be cautious about ambiguous cases (where two possible triangles might satisfy the given conditions).

Example:

Suppose you have:

  • A=40A = 40^\circ
  • B=60B = 60^\circ
  • a=8a = 8

First, find the third angle: C=180AB=1804060=80C = 180^\circ - A - B = 180^\circ - 40^\circ - 60^\circ = 80^\circ

Now, apply the Law of Sines to find side bb:

sinAa=sinBb\frac{\sin A}{a} = \frac{\sin B}{b}

Substitute the known values: sin408=sin60b\frac{\sin 40^\circ}{8} = \frac{\sin 60^\circ}{b}

Solve for bb: b=8sin60sin40b = \frac{8 \cdot \sin 60^\circ}{\sin 40^\circ} b80.8660.64310.78b \approx \frac{8 \cdot 0.866}{0.643} \approx 10.78

Therefore, side b10.78b \approx 10.78.

Let me know if you'd like further details or explanations!

Here are 5 related questions to explore further:

  1. How does the Law of Sines work for obtuse triangles?
  2. What is the ambiguous case (SSA) in triangle solving?
  3. How does the Law of Cosines complement the Law of Sines?
  4. Can the Law of Sines be used for non-right triangles?
  5. What is the difference between solving triangles using the Law of Sines and the Law of Cosines?

Tip: Always check for the ambiguous case when using the Law of Sines with two sides and a non-included angle.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Triangle Geometry
Law of Sines

Formulas

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

Theorems

Law of Sines
Triangle Angle Sum Theorem

Suitable Grade Level

Grades 9-11