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Tana * tanb is known in triangle ABC
tanA*tanB90
The triangle ABC is an obtuse triangle
(2) When cosacosb > 0
sinAsinB0
cos(A+B)>0
A+B90
Combining (1) and (2), the triangle ABC is an obtuse triangle
If a = {A2, a + 1, - 1}, B = {2a-1, | A-2 | 3a2 + 4}, and a ∩ B = {- 1}, then a=______ ．
From a ∩ B = {- 1}, we get that - 1 belongs to both set a and set B, and | A-2 | ≥ 0, 3a2 + 4 ≥ 0, then 2a-1 = - 1, so a = 0, then a = {0, 1, - 1}, B = {- 1, 2, 4} satisfies the meaning of the problem, so the answer is: 0
Given the set a = {2, a2-a + 3, A2 + 2A + 3}, B = {1, A-3, A2 + A-4, a2-3a + 7}, and a ∩ B = {2, 5}, find the value of real number a
A = {2, a2-a + 3, a2-a + 3, A2 + 2 + 2A + 3}, B = {1, A-3, A2 + A-4, a2-3a + 7}, and a ∩ B = {2,5,5}, {5 ∩ B = {2,5,5}, a = {2,23,38}, but 5 ∉ a, and a ∉ a, and a ∩ B = {2,5,5, B = {2,5,5,5,5,5,5,5}; if a + A-4 = 2, that is a ? 2, a ? B = {B = {2, B = {2,5 whena = 2, a = {2, 5, 11}, B = {1, - 1, 2, 5}, the condition is satisfied; If a2-3a + 7 = 2, that is, a2-3a + 5 = 0, △ = 9-20 = - 11 < 0, the equation has no solution. To sum up, a = 2
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