In a triangular pyramid a-bcd, plane abd ⊥ plane DBC, ∠ ABC = ∠ DBC = 120 ° AB = BC = BD, the tangent of plane angle of dihedral angle a-bd-c is obtained Urgent ah, homework, ask experts to help

In a triangular pyramid a-bcd, plane abd ⊥ plane DBC, ∠ ABC = ∠ DBC = 120 ° AB = BC = BD, the tangent of plane angle of dihedral angle a-bd-c is obtained Urgent ah, homework, ask experts to help

To tell you clearly, there is no such d point
D is point B, that is to say, plane abd ⊥ plane DBC and ∠ ABC = ∠ DBC = 120 ° cannot hold at the same time

If plane abd ⊥ plane BDC, △ BCD is an equilateral triangle, ab = ad, ∠ bad = 90 °, then the tangent value of dihedral angle A-CD - is

If plane abd ⊥ plane BDC, △ BCD is an equilateral triangle, ab = ad, ∠ bad = 90 °, then the tangent value of dihedral angle a-cd-b is
Make AE ⊥ DB to e, make ef ⊥ DC to F, connect AF
∵ planar abd ⊥ planar BDC
⊥ AE ⊥ plane BDC
∴AE⊥DC
⊥ CD ⊥ plane AEF
∴CD⊥AF
The angle of AFE is a-cd-b
∵AB=AD,∠BAD=90°
Ψ△ abd is an isosceles right triangle
Then: AE = de
tan∠AFE=AE/EF=DE/EF=1/EF/DE=1/sin∠D=1/sin60°=2√3/3

Through the circle X & # 178; + Y & # 178; = R & # 178; interior point P0 (x0, Y0) lead chord AB, take a, B as the tangent point of two tangent intersection point as P, find the trajectory equation of P

Let P (x, y)
Then the slope of line OP is K1 = Y0 / x0 = Y / X
OP is perpendicular to AB, and the slope of line AB is K2 = - 1 / K1 = - x0 / Y0 = - X / y
Straight line AB passes through point P0: y-y0 = (- X / y) * (x-x0), that is, x ^ 2-x0 * x + y ^ 2-y0 * y = 0 → (x-x0 / 2) & # 178; + (y-y0 / 2) & # 178; = [(x0) & # 178; + (Y0) & # 178;] / 4
The trajectory equation of P is a circle

What is the tangent equation passing through a point P (2,8 / 3) on the cubic power of the curve y = (1 / 3) x?
Please write down the solution and detailed steps, I just learned derivative,

The tangent equation is Y-8 / 3 = 4 (X-2) and y = 4x-16 / 3 is reduced to y = 4x-16 / 3. First, verify whether the given point is on the curve, then derive the tangent slope from the given point, then write the equation obliquely, and finally simplify it to a general equation. If the point is not on the curve, then set the tangent curve

High school mathematics garden equation and tangent equation, welcome to challenge
The straight line y = x + B divides the circle x ^ 2 + y ^ 2-8x + 2Y + 8 = 0 into two arcs of equal length, then B =?
Tangent equation of function y = (x ^ 2) / 4 at point P (2,1)

The straight line y = x + B divides the circle x ^ 2 + y ^ 2-8x + 2Y + 8 = 0 into two arcs of equal length, then B = (x-4) ^ 2 + (y + 1) ^ 2 = 9 the coordinates of the center of the circle (4, - 1) the straight line divides the circle into two equal arcs, indicating that the straight line passes through the center of the circle

The equation of circle
There is a moving point P (T, t), a moving point m on the circle x ^ 2 + (Y-1) ^ 2 = 1 / 4, and a moving point n on the circle (X-2) ^ 2 + y ^ 2 = 1 / 4. What is the maximum value of PN pm?

When p is (0,0) is the maximum, M is (0,1 / 2) and N is (5 / 2,0), the maximum is 2. If the first circle is symmetric with respect to the straight line y = x, it can be found that it is tangent to another circle. If we want to take a point on the two circles respectively, we know that the farthest point on the two tangent circles is the sum of the diameters of the two circles

Let point C be any point of the curve y = 2x (x > 0), and the circle with point C as the center intersects with X axis at points E and a, and intersects with y axis at points E and B. (1) prove that the area of polygon eacb is a fixed value, and find this fixed value; (2) let the line y = - 2x + 4 intersect with circle C at points m and N, if | EM | = | en |, find the equation of circle C

(1) It is proved that: point C (T, 2t) (T > 0), because the circle with point C as the center intersects with X axis at points E and a, and intersects with y axis at points E and B. so point E is the origin of rectangular coordinate system, that is e (0, 0). Then the equation of circle C is (x − T) 2 + (Y − 2t) 2 = T2 + 4T2. Then a (2t, 0), B (0, 4T). By | CE | = | Ca | = | CB

As shown in the figure, the circumscribed quadrilateral ABCD of ⊙ o is a right angled trapezoid, ad ∥ BC, ∠ a = ∠ B = 90 degrees. (1) try to explain OC ⊥ OD; (2) if CD = 4cm, ∠ BCD = 60 degrees, find the radius of ⊙ o

(1) As shown in the figure, connect OD, ∵ ad ∥ BC, ∵ BCD + ∠ ADC = 180 °, ∵ ODC = 12 ∠ ADC, ∵ OCD = 12 ∠ BCD, ∵ ODC + ∠ OCD = 12 ∠ ADC + 12 ∠ BCD = 90 ° and ⊥ OC ⊥ OD; (2) through D as de ⊥ BC in E, abed is a rectangle, and De is equal to the diameter of ⊙ O. in RT △ Dec, ∵ Dec = 90 °, ∵ ECD = 60 °, CD = 4cm, ∵ CE = 12CD = 2cm, de = CD2 − CE2 = 23cm

It is helpful for the responder to give an accurate answer, as shown in the figure, in the rectangular trapezoid ABCD, ab = 7, ∠ B = 90 ° BC-AD = 1, the circle with diameter CD intersects AB at two points e, F, and AE = 1, whether there is a point P on the line AB, so that the triangle with vertices P, a, D is similar to the triangle with vertices P, B, C? If not, explain the reason; if there are several such points? And calculate the length of AP

The length of AP is 50

Given the rectangular trapezoid ABCD, ad parallel BC e is the hypotenuse, and the midpoint of CD is made EF ⊥ AB, it is proved that EF is the median line of rectangular trapezoid ABCD

It is proved that if DF is connected and extended, and the extension line of CB is at point G