In the triangle ABC, AB * BC = 3, and the area of the triangle ABC is between √ 3 / 2 and 3 / 2, then the value range of the angle between AB and BC AB and BC are vectors

In the triangle ABC, AB * BC = 3, and the area of the triangle ABC is between √ 3 / 2 and 3 / 2, then the value range of the angle between AB and BC AB and BC are vectors

From S = ab · BC · SINB / 2
=（3/2）sinB,
∴√3/2≤3sinB/2≤3/2,
√3/3≤sinB≤1.
From SINB = √ 3 / 3,
∴∠B=35°,
When ∠ B ＞ 0,
∠B=35+90=125°,
∴35°≤∠B≤125°.

Using elimination method to find 2x + 3Y = 0, 3x + 5Y = - 1,

If 2x + 3Y = 0, then x = - 3 / 2Y is substituted by 3x + 5Y = - 1
-9/2y+5y=-1
1/2y=-1
Then y = - 2
Then x = - 3 / 2Y = 3

If 1 / X-1 / y = 3, find 5x + 2xy-5y of x-4xy-y

1/x-1/y=3
∴y-x=3xy
Primitive = [(X-Y) - 4xy] / [5 (X-Y) + 2XY]
=(-3xy-4xy)/(-15xy+2xy)=(-7xy)/(-13xy)
=7/13