If the inclination angle α ∈ [π / 4, π / 2) ∪ (π / 2,3 π / 4] in the straight line, then the value range of slope k? That's right

If the inclination angle α ∈ [π / 4, π / 2) ∪ (π / 2,3 π / 4] in the straight line, then the value range of slope k? That's right

From the image of k = Tan α and tangent function, we can see that α ∈ (- ∞, - 1] ∪ [1, + ∞)

In (0,2 π), find the range of α which makes sin α * cos α 0 hold at the same time

When a = 3 π / 4, the sum of the two is 0, when (3 π / 4, π) | Sina | < | Cosa |, it does not match, when (3 π / 4, π / 2) | Sina | > | Cosa |, it is consistent with that cosa in the fourth quadrant is positive, when (7 π / 4, 2 π) |

The sine of the inclination angle α of a straight line is equal to 3 / 5, and the slope k of the straight line is?

It is known that sina = 3 / 5
So cosa = ± 4 / 5
k=tana=±3/4

The range of the slope of the line is r, right?

Yeah

If it is known that △ ABC and △ def are similar and the ratio of the corresponding median line is 2:3, then the perimeter ratio of △ ABC and △ DEF is______ ．

∵△ ABC and △ def are similar, and the ratio of the corresponding median line is 2:3, and the similarity ratio of them is 2:3; therefore, the perimeter ratio of △ ABC and △ DEF is 2:3

If the length of three sides of triangle ABC is 3:4:5 and the perimeter of other similar triangle DEF is 18, the area of DEF is

13.5

If the area ratio of triangle ABC and triangle DEF is 9:16, the perimeter of triangle ABC and triangle DEF is

Because the area ratio is equal to the square of the similarity ratio = 9:16
So the similarity ratio is 3:4
Perimeter of triangle ABC: perimeter of triangle def = 3:4

In △ ABC and △ def, ab = 2DE, AC = 2DF, ∠ a = ∠ D. if the perimeter of △ ABC is 16 and the area is 12, then the perimeter and area of △ def are ()
A. 8，3B. 8，6C. 4，3D. 4，6

Because in △ ABC and △ def, ab = 2DE, AC = 2DF, ∵ ABDE = ACDF = 2, and ∵ a = ∠ D, ∵ ABC ∽ def, and the similarity ratio of △ ABC and △ DEF is 2, the perimeter of ∵ ABC is 16, the area is 12, and the perimeter of ∵ DEF is 16 △ 2 = 8, the area is 12 △ 4 = 3

In △ ABC and △ def, ab = 2DE, AC = 2DF, ∠ a = ∠ D. if the perimeter of △ ABC is 16 and the area is 12, then the perimeter and area of △ def are ()
A. 8，3B. 8，6C. 4，3D. 4，6

Because in △ ABC and △ def, ab = 2DE, AC = 2DF, ∵ ABDE = ACDF = 2, and ∵ a = ∠ D, ∵ ABC ∽ def, and the similarity ratio of △ ABC and △ DEF is 2, the perimeter of ∵ ABC is 16, the area is 12, and the perimeter of ∵ DEF is 16 △ 2 = 8, the area is 12 △ 4 = 3

It is known that the triangle ABC is similar to the triangle def, and De is 3cm, AB is 4cm, hold is 5cm, CA is 6cm. The perimeter of the triangle DEF is calculated
The answer I know is that I don't know how to write the standard process

de/ab =ef/bc =df/ac
3/4=ef/5,3/4=df/6
ef=15/4,df=18/4
Def perimeter = 15 / 4 + 18 / 4 + 3 = 11.25