If ABC = 1, then the value of AAB + A + 1 + BBC + B + 1 + CCA + C + 1 is () A. 1B. 0C. -1D. -2

If ABC = 1, then the value of AAB + A + 1 + BBC + B + 1 + CCA + C + 1 is () A. 1B. 0C. -1D. -2

Then AAB + A + 1 + BBC + B + 1 + CCA + C + 1 = AC1 + AC + C + BBC + B + 1 + BC1 + BC + B = ABCB + 1 + BC + BB + B + 1 + BC1 + BC + B = 1 + B + BCB + 1 + BC = 1

In △ ABC, the length of BC can be obtained by ∠ B = 60 °, C = 45 ° AB + AC = 2 + √ 6

Let BD be x, then according to the Pythagorean theorem of right triangle, ab = 2x, ad = BC = √ 3x, AC = √ 6x, AB + AC = = 2x + √ 6x = 2 + √ 6, we deduce x = 1, so BC = 1 + + 3

In △ ABC, ∠ B = π 6, AC = 1, ab = 3, then the length of BC is______ ．

Because in △ ABC, ∠ B = π 6, AC = 1, ab = 3, so sinc = 3 × sinπ 6 = 32 is tested by sine theorem. So C = π 3 or C = 2 π 3, when C = π 3, triangle is right triangle, so BC = 2, when C = 2 π 3, triangle is isosceles triangle, so BC = 1, so the answer is: 1 or 2

It is known that in △ ABC, ab = 5, BC = 2A + 1, AC = 12

In ∵ △ ABC, ab = 5, BC = 2A + 1, AC = 12, ∵ 2A + 1 ＞ 12 − 52A + 1 ＜ 12 + 5, the solution is 3 ＜ a ＜ 8, so the range of a is 3 ＜ a ＜ 8