Answer:
Explanation:
To find the missing side length and angles of triangle ABC, given that angle C is 129°, side a is 7 units, and side b is 10 units, we can use the Law of Cosines and the Law of Sines.
Finding side c using the Law of Cosines:
According to the Law of Cosines, in a triangle ABC, c^2 = a^2 + b^2 - 2abcos(C).
Substituting the given values: c^2 = 7^2 + 10^2 - 2(7)(10)cos(129°).
Calculating: c^2 ≈ 49 + 100 + 140cos(129°).
Simplifying: c^2 ≈ 149 + 140(-0.64278760968).
Calculating: c^2 ≈ 149 - 89.99852275552.
Therefore, c^2 ≈ 59.00147724448.
Taking the square root: c ≈ √59.00147724448.
Hence, c ≈ 7.68 units (rounded to two decimal places).
Finding angle A using the Law of Sines:
According to the Law of Sines, sin(A)/a = sin(C)/c.
Substituting the given values: sin(A)/7 = sin(129°)/7.68.
Cross-multiplying: 7sin(A) = 7.68sin(129°).
Calculating: sin(A) ≈ (7.68sin(129°))/7.
Taking the inverse sine: A ≈ sin^(-1)((7.68sin(129°))/7).
Hence, A ≈ 49.3° (rounded to one decimal place).
Finding angle B:
Since the sum of the angles in a triangle is 180°, angle B = 180° - angle A - angle C.
Substituting the given and calculated values: B = 180° - 49.3° - 129°.
Calculating: B ≈ 1.7° (rounded to one decimal place).
Therefore, in triangle ABC:
Side a = 7 units
Side b = 10 units
Side c ≈ 7.68 units (rounded to two decimal places)
Angle A ≈ 49.3° (rounded to one decimal place)
Angle B ≈ 1.7° (rounded to one decimal place)
Angle C = 129°