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The waist length of isosceles triangle is 10cm, the area is 3cm2 under 25 heel trigonometric function
Let the bottom edge be 2x, then the height is √ (100-x & # 178;)
x·√(100-x²)=25√3
The solution is X & # 178; = 50 ± 25
When X1 = 5 √ 3, half of the vertex angle is arcsin [(5 √ 3) / 10] = 60 degree
When x2 = 5, half of the vertex angle is arcsin (5 / 10) = 30 degrees
So when the bottom edge is 10 √ 3cm, the vertex angle is 120 degrees; when the bottom edge is 10cm, the vertex angle is 60 degrees
If the plane formed by positive △ ABC and positive △ BCD is vertical, then the dihedral angle a-bd-c sine value should be calculated by vector method!
Take e in BC, make ef ⊥ BD in F, connect AE, AF
The planes of ABC and BCD are perpendicular to each other,
⊥ AE ⊥ plane BCD,
AF⊥BD,
The plane angle of the dihedral angle a-bd-c,
AE=(√3/2）BC,EF=(√3/4）BC,
tan∠AFE=AE/EF=2,
The calculated value is: sin ∠ AFE = (2 √ 5) / 5
If the plane formed by positive △ ABC and positive △ BCD is perpendicular, then the sine value of dihedral angle a-bd-c is How to do it with vector method?
Let B (1,0,0), then d (0, √ 3,0), a (0,0, √ 3), vector Ba = (- 1,0, √ 3), BD = (- 1, √ 3,0), normal vector of plane abd be n = (P, Q, 1), then n * Ba = - P + √ 3 = 0, n * BD = - P +
In the triangular pyramid a-bcd, the bottom is an isosceles right triangle, BC = CD, AB is perpendicular to the plane BCD, and ab = BC
Let AB = BC = CD = a
∵ ab ⊥ plane BCD
∴AB⊥BD AB⊥CD AB⊥BC
∴AC=(AB^2+BC^2)^1/2=√2a
∵ △ BCD is isosceles right triangle, and BC = CD
∴BD=(BC^2+CD^2)^1/2=√2a
∴AD=(AB^2+BD^2)^1/2=√3a
∵CD⊥BC CD⊥AB
⊥ CD ⊥ ABC
∴CD⊥AC
That is, the angle between AD and plane ABC
sin∠DAC=CD/AD=√3/3
If a tangent is introduced from point P (2,3) to circle (x-1) 2 + (Y-1) 2 = 1, then the tangent equation is______ ．
(1) If the slope of the tangent exists, let the equation of the tangent be Y-3 = K (X-2), that is, kx-y-2k + 3 = 0, then the distance from the center of the circle to the tangent d = | K − 1 − 2K + 3 | K2 + 1 = 1, and the solution is k = 34, so the equation of the tangent is 3x-4y + 6 = 0 (2)
Two points P and Q on the curve X ^ 2 + y ^ 2 + x-6y + 3 = 0 satisfy: (1) symmetry about the straight line kx-y + 4 = 0; (2) op vertical OQ
Two points P and Q on the curve X ^ 2 + y ^ 2 + x-6y + 3 = 0 satisfy the following conditions: (1) the straight line kx-y + 4 = 0 is symmetric; (2) O is the origin, OP is perpendicular to OQ
Find the equation of line PQ
Note: O is the origin
The curve X ^ 2 + y ^ 2 + x-6y + 3 = 0 is a circle, the standard equation is: (x + 1 / 2) ^ 2 + (Y-3) ^ 2 = 25 / 4, the center of the circle (- 1 / 2,3) radius is 5 / 2, the straight line kx-y + 4 = 0 passes through the point (0,4), then the distance from the point to the point P and the point q is equal; in addition, the distance from the center of the circle to the point P and the point q is also equal, so the line between the point (0,4) and the center of the circle is the vertical bisector of PQ, oblique
It is known that two points P and Q on the circle x + y + x-6y + 3 = 0 satisfy the following conditions: (1) they are symmetric with respect to the straight line kx-y + 4 = 0; (2) op ⊥ OQ (o is the center of the circle)
Firstly, the curve equation is changed to: (x + 1 / 2) ^ 2 + (Y-3) ^ 2 = (5 / 2) ^ 2. If this is a circle, then PQ is on the circle and PQ is symmetrical with respect to the straight line. Then this straight line is the vertical bisector of the line PQ. The straight line must pass through the center of the circle (- 1 / 2,3) and substitute the center of the circle on the straight line to get - K / 2 - 3 + 4 = 0, k = 2, and the straight line is 2x-y + 4 = 0 (1)
There are two points P on the curve X square + y square + x-6y = 0, Q satisfies the equation that PQ is symmetric about the line kx-y + 4 = 0, and OP is perpendicular to OQ
From x square + y square + x-6y = 0, we get: (x + 1 / 2) ^ 2 + (Y-3) ^ 2 = 37 / 4, because P and Q are points on the circle, and P and Q satisfy the symmetry of PQ about the line kx-y + 4 = 0, so the line kx-y + 4 = 0 passes through the center of the circle (- 1 / 2,3) and is substituted into the linear equation: - 1 / 2 * K-3 + 4 = 0, and the solution is: k = 2
If a point p outside the circle (X-2) ^ 2 + y ^ 2 = 2 is taken as two mutually perpendicular tangents of the circle, then the trajectory equation of the moving point P is obtained
Circle O: (X-2) ^ 2 + y ^ 2 = 2, O (2,0), R ^ 2 = 2
Two mutually perpendicular tangents PA ⊥ Pb, then OA ⊥ PA, ob ⊥ Pb
PA=PB=OA=OB=√2,OP=2
P(x,y)
The trajectory equation circle of the moving point P: (X-2) ^ 2 + y ^ 2 = 4
Find the trajectory equation of the center of a circle on a straight line and passing through the intersection of two circles. The data are as follows
Find the trajectory equation of the intersection of two circles x ^ 2 + y ^ 2-4x-3 = 0 and x ^ 2 + y ^ 2-4y-3 = 0 with the center of the circle on the straight line X-Y-4 = 0
The equation of this circle can be written as: x ^ 2 + y ^ 2-4x-3 + K (x ^ 2 + y ^ 2-4y-3) = 0, that is, (1 + k) x ^ 2-4x + (1 + k) y ^ 2-4ky-3-3k = 0. The reason why this circle is a circle is that the coefficients of the square terms of X and y are the same, so it can always be formulated as a circle
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