Question

数学题，初一下学期的~~

在△ABC中，∠B：∠C=3:4，∠A-∠B=10°，则∠A=多少度？？

Answer 1

∠B = ∠A - 10°
∠C = 4∠B/3 = (4/3)(∠A - 10°)

∠A + ∠B + ∠C = 180°
∠A + (∠A - 10°) + (4/3)(∠A -10°) = 180°
∠A + (7/3)(∠A - 10°) = 180°
3∠A + 7(∠A - 10°) = 540°
10∠A = 540°+ 70°
∠A = 61°

∠B = 51°
∠C = 68°

Answer 2

61

Answer 3

∠B：∠C=3:4
3∠C=4∠B
∠C=(4/3)∠B
∠A-∠B=10°
∠A=10°+∠B
∠A+∠B+∠C=180°
10°+∠B+∠B+(4/3)∠B=180°
(10/3)∠B=180-10
(10/3)∠B=170°
∠B=51°
所以∠A=10°+∠B=10+51=61°