In the acute triangle ABC, AB > BC > AC, and the maximum internal angle is 24 ° larger than the minimum internal angle, then the value range of ∠ A is______ ．

In the acute triangle ABC, AB > BC > AC, and the maximum internal angle is 24 ° larger than the minimum internal angle, then the value range of ∠ A is______ ．

Let ∠ B = x, then ∠ C = x + 24 °, ∵ AB > BC > AC, ∵ C > a > b, ∵ a = 180 ° - 24 ° - 2x = 156 ° - 2x, ∵ x + 24 ° > 156 °− 2x > XX + 24 °＜ 90 °, ∵ 44 °＜ x ＜ 52 °, ∵ 52 °＜ a ＜ 68 °. So the answer is 52 °＜ a ＜ 68 °

In the acute triangle ABC, AB > BC > AC, and the maximum internal angle is 24 ° larger than the minimum internal angle, then the value range of ∠ A is______ ．

Let ∠ B = x, then ∠ C = x + 24 °, ∵ AB > BC > AC, ∵ C > a > b, ∵ a = 180 ° - 24 ° - 2x = 156 ° - 2x, ∵ x + 24 ° > 156 °− 2x > XX + 24 °＜ 90 °, ∵ 44 °＜ x ＜ 52 °, ∵ 52 °＜ a ＜ 68 °. So the answer is 52 °＜ a ＜ 68 °

In the acute triangle ABC, AB > BC > AC, and the maximum internal angle is 24 ° larger than the minimum internal angle, then the value range of ∠ A is______ ．

Let ∠ B = x, then ∠ C = x + 24 °, ∵ AB > BC > AC, ∵ C > a > b, ∵ a = 180 ° - 24 ° - 2x = 156 ° - 2x, ∵ x + 24 ° > 156 °− 2x > XX + 24 °＜ 90 °, ∵ 44 °＜ x ＜ 52 °, ∵ 52 °＜ a ＜ 68 °. So the answer is 52 °＜ a ＜ 68 °

In the triangle ABC, the edges corresponding to the angles a, B and C are ABC respectively. It is known that 4acosa + ccosb + bcosc = 0. (1) find the value of cosa. (2) a = 4, B + C = 5,
(2) A = 4, B + C = 5, find the area of vector AB * vector AC and triangle ABC

(1) Sine theorem 3sinacosa = sinccosb + sinbcosc = sin (B + C) = Sina
Then 3cosa = 1,
So cosa = 1 / 3
(2) Cosine theorem a = B + c-2bccosa = (B + C) - 2bc-2bccosa
Substituting a = 4, B + C = 5, cosa = 1 / 3, BC = 27 / 8
bccosA=9/8
That is: vector AB * vector AC = 9 / 8
S = 1 / 2bcsina = 9 times radical 2 / 8