What does it mean that the product of two slopes is 1 (reciprocal each other)? ditto

What does it mean that the product of two slopes is 1 (reciprocal each other)? ditto

Product - 1 is vertical, equality is parallel, if it is 1, it is a general intersection (positive proportion function is inverse function)

The product of the slope of a straight line formed by B (a, 0) and C is K. k = - 2
Point a (- A, 0) point B (a, 0) and point C form a straight line. The product of the slope is K. if k = - 2, then the triangle ABC is an acute triangle

1. First of all, it can be seen from the drawing that angles a and B must be acute (no statement)
2. It only needs to solve that the angle c is an acute angle, that is, cos (c) > 0, cos (180 - (a + b)) > 0
Because: from the slope product not - 2, Tan (a) * Tan (180-b) = - 2 = = > Tan (a) * Tan (b) = 2 = = > sin (a) * sin (b) / cos (a) * cos (b) = 2 > 0
We can also get: sin (a) * sin (b) = 2cos (a) * cos (b)
Because: cos (180 - (a + b)) = - cos (a + b) = sin (a) * sin (b) - cos (a) * cos (b) = cos (a) * cos (b)
If there is 1, both a and B are acute angles, then cos (a) * cos (b) > 0
Thus: cos (180 - (a + b)) = cos (a) * cos (b) > 0 = = > cos (c) > 0
That is, angle c is also an acute angle

It is proved that the slope of the straight line is greater than zero
The function f (x) = loga (a is the base) (a ^ x-1) (a > 1) is known. If a (x1, Y1), B (X2, Y2) are two different points on the image of F (x), it is proved that the slope of line AB is greater than zero

Suppose X1 > x2 domain a ^ X-1 > 0 a ^ x > 1, that is, a ^ x > A ^ 0 a > 1, so a ^ x is an increasing function, so x > 1, so X1 > x2 > 0, y1-y2 = f (x1) - f (x2) = loga [(a ^ x1-1) / (a ^ x2-1)] a > 1, so a ^ x is an increasing function, so a ^ X1 > A ^ x2 and a ^ x > 1, so a ^ x1-1 > A ^ x2-1 = 0, so (a ^ x1-1)

How to prove that the product of the vertical slopes of two lines is one? Detailed method is needed!

Let the slope of two straight lines be K1, K2, and the inclination angle be a, B. If two straight lines are vertical, then the angle between them is 90 degrees, so tan (a-b) = tan90 = (Tana tanb) / (1 + tanatanb) = infinity, because Tana = K1, tanb = K2, so 1 + tanatanb = 1 + k1k2 = 0, so k1k1 = - 1

As shown in the figure, in the right angle △ ABC, ∠ C = 90 ° and ∠ ABC = 2 ∠ a, BD is the bisector of the angle, and the degrees of ∠ A and ∠ CDB are calculated

The picture is in (=@__ @=) where?

In the RT triangle ABC, angle c = 90 degrees, angle a = 45 degrees, if AB = 6cm, then BC=

In RT triangle ABC, if angle c = 90 degrees, angle a = 45 degrees, then angle B = 45 degrees, AC = BC
Let BC = x, then AC = BC = x, the square of X + the square of x = 6
2X squared = 36
Square of x = 18
X = radical 18 = 3 radical 2

In △ ABC, the edges of angles a, B and C are a, B and C respectively, and a = 2, CoSb = 45. (1) if B = 3, find the value of sina; (2) if the area of △ ABC is s △ ABC = 3, find the value of B and C

(1) Because CoSb = 45 and 0 ＜ B ＜ π, SINB = 1 − cos2b = 35. From the sine theorem, Sina = asinbb = 25. (2) because s △ ABC = 12acsinb = 3, so 12 × 2C × 35 = 3. So C = 5. From the cosine theorem, B2 = A2 + C2 − 2accosb = 22 + 52 − 2 ×× 2 × 5 × 45 = 13. So B = 13

In the triangle ABC, a, B and C are the opposite sides of the three angles a, B and C respectively. If a = 2, C = π / 4, CoSb / 2 = 2, and 5 / of the sign, we can find the C side
In the triangle ABC, a, B and C are the opposite sides of the three angles a, B and C respectively. If a = 2, C = π / 4, CoSb / 2 = 2 and 5 / 5, the length of side C can be obtained

cosB=2(cosB/2)^2-1=8/5-1=3/5,
sinB=4/5,
According to the sine theorem,
AB/sinC=a/sinA,
c=(2sinπ/4)sin[π-B-C]
=√2(sinBcosC+cosBsinC)
=√2[(4/5)*√2/2+(3/5)*√2/2]
=7/5.
The length of side C is 7 / 5

The circumference of the triangle is 32 cm, and there is a point O in it. The distance from the point O to the three sides is 4 cm. Find the face of the triangle ABC

Let the triangle be ABC
AB + AC + BC = 32
Triangle area = AB * 4 / 2 + AC * 4 / 2 + BC * 4 / 2 = (AB + AC + BC) * 2 = 64

In the triangle ABC, there is a point O. the distance from the point O to the three sides is 2 cm. We also know that the circumference of the triangle ABC is 20 cm

A: according to the meaning of the question, this point O is the center of the inscribed circle of the triangle ABC, r = 2cm;
Connect Ao, Bo, Co, triangle ABC area:
S = s triangle ABO + s triangle BCO + s triangle ACO
=AB*R/2+BC*R/2+AC*R/2
=(AB+BC+AC)*R/2
=20*2/2
=20 cm²