Solve the triangle α = 27 degrees 30', c = 52.8, a = 28.1
sinC/c = sinA/a, so
sinC/52.8 = sin27°30'/28.1
sinC = 0.8676
C = 60°11' or 119°49'
Since A+C < 180 in either case, there are two possible triangles.
Knowing A and C, you now know what B can be, and then use the law of sines again to find b.
To solve triangle α = 27 degrees 30', c = 52.8, and a = 28.1, we can use the law of sines and the law of cosines.
1. Law of Sines:
The law of sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant. In this case, we have:
sin(α) / a = sin(β) / b = sin(γ) / c
2. Law of Cosines:
The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. In this case, we have:
c² = a² + b² - 2ab * cos(γ)
Let's use these equations to solve the triangle:
1. Find angle β using the law of sines:
sin(α) / a = sin(β) / b
sin(27.5°) / 28.1 = sin(β) / b
To find sin(β), multiply both sides by b:
b * sin(27.5°) / 28.1 = sin(β)
Now, solve for β by taking the inverse sine of both sides:
β = arcsin(b * sin(27.5°) / 28.1)
2. Find angle γ:
Since we know two angles in a triangle, we can find the third angle by subtracting the sum of the known angles from 180 degrees:
γ = 180° - α - β
γ = 180° - 27.5° - β
3. Now, we can use the law of cosines to find side b:
c² = a² + b² - 2ab * cos(γ)
Substituting the given values:
52.8² = 28.1² + b² - 2 * 28.1 * b * cos(γ)
Now, substitute the value of γ and solve for b:
52.8² - 28.1² = b² - 2 * 28.1 * b * cos(180° - 27.5° - β)
b² - 2 * 28.1 * b * cos(152.5° - β) = 52.8² - 28.1²
Solve for b using a numerical method or a graphing calculator, as this equation is not solvable algebraically.
Once you find the value of b, you can substitute it back into the law of sines to find the remaining angles β and γ, and have all the information about the triangle.