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If a triangle with three sides BC = 3, AC = 4, ab = 5 is folded into △ ABC 'along the longest side AB, then the length of CC' is () A. 125B. 512C. 245D. 524
∵ BC = 3, AC = 4, ab = 5 ∵ ABC is a right triangle. The length of ∵ CC 'is equal to twice the height of the hypotenuse of ∵ ABC. Suppose the height of the hypotenuse is h. according to the area of ∵ ABC = 12bc · AC = 12ab · h, the length of H = 125 ∵ CC' is 245
If a triangle with three sides BC = 3, AC = 4, ab = 5 is folded into △ ABC 'along the longest QB, then the length of CC' is ()
The cosine of angle c is equal to 0.6
Using the formula of double angle of cosine
The cosine value of the angle c of 2 times is equal to -0.28
Then we use cosine formula to find the cosine value of double angle C
The solution is C = 4.8
In the triangle ABC, BC = 3, AC = 4, ab = 5, the triangle ABC "is obtained by folding the triangle along the longest side AB, then the length of CC" is equal to
From Pythagorean theorem, we know that △ ABC is a right triangle and ∠ C = 90 degree
AC = AC ", obviously CC" ⊥ ab
From the equal area of △ ABC, it can be concluded that
AC*BC/2=AB*CC"/2/2
That is CC "= 2Ac * BC / AB = 2 * 4 * 3 / 5 = 24 / 5
In a dihedral angle of 60 degrees, there is a point whose distance to another plane is cm. How about the distance to the edge?
Let the distance between a point on the plane and another plane be X
Make another plane through point a and perpendicular to the intersection line L. the perpendicular feet are B and C respectively
Connect BC, then ab ⊥ BC, ∠ ACB = 60 degree
BC = x / tan60 ° = (radical 3) x / 3
The distance to the other edge is (root 3) x / 3
From the point P (m, 3) to the circle (x + 2) ^ 2 + (y + 2) ^ 2 = 1, the tangent line is introduced, and the equation to make the length of the tangent line the shortest tangent line is obtained
(length of tangent) ^ 2 = (distance from P to the center of circle) ^ 2-r ^ 2 make the length of tangent the shortest, then the shortest point P (m, 3) from P to the center of circle is on the straight line y = 3, and the shortest distance is the distance from the center of circle to the straight line = 5. At this time, M = - 2 point P (- 2,3) let the tangent slope be KY-3 = K (x + 2) kx-y + 3 + 2K = 0, and the distance from the center of circle is equal to the radius R1 = | - 2K
The tangent equation of a circle made by a point (2.1) over the circle x + y = 5
And where are the knowledge points (such as circle or derivative)
Since it is a tangent, the tangent must be perpendicular to the radius passing through the tangent
So first find the slope of the radius, and then find the tangent slope. If the slope is known and the point is known, the equation will come out
The slope of radius is k = (1-0) / (2-0) = 1 / 2
Then: the slope of tangent is: K '= - 2 (the product of slopes of two lines perpendicular to each other is - 1)
So the tangent equation is: Y-1 = - 2 (X-2)
That is: y = - 2x + 5
Given that the line y = MX + 4 is tangent to the circle x + y = 4, find the value of M and tangent equation
Center (0,0)
Distance from center of circle to line = radius
r=2
Using the formula of distance from point to line
2 = 4 / (radical m ^ 2 + 1)
m^2 =b
M = radical 3 or - radical 3
Y = radical 3 + 4 or y = - radical 3x + 4
The tangent equation of a circle passing through a point m (1,2) on the circle x2 + y2 = 5 is______ ．
From the circle x2 + y2 = 5, the coordinates of the center a are (0, 0), the radius of the circle r = 5, and | am | = 5 = R, so if M is on the circle, the tangent of the circle passing through M is perpendicular to the line where am is located, and m (1, 2), the slope of the line where am is located is 2, so the slope of the tangent is - 12, then the tangent path is Y-2 = - 12 (x-1), that is, x + 2y-5 = 0. So the answer is: x + 2y-5 = 0
Circle C (x-1) ^ 2 + y ^ 2 = 2, solved the tangent equation of M (2,3). Quick, I got one X = 2 and one y = 8 / 15x + 61 / 15,
Is (x-1) ^ 2 + y ^ 2 = 1. The radius is 1
Answer: circle C: (x-1) ² + Y & #178; = 2 center of circle (1,0), radius r = √ 2, let the straight line passing through point m (2,3) be Y-3 = K (X-2), kx-y-2k + 3 = 0, the straight line is tangent to the circle, and the distance from the center of circle to the straight line d = R, so: D = | k-0-2k + 3 | / √ (k ^ 2 + 1) = √ 2, so: | 3-K | = √ (2k ^ 2 + 2) the square of both sides is: K ^ 2-6k + 9 =
Through the point m (2.3) to the circle (x-3) + (y + 1) = 1, the tangent line is drawn, and the tangent equation is solved
The center coordinate of the circle is (3, - 1) and the radius is 1. Then the line x = 2 is a tangent of the circle
Let the other tangent equation be Y-3 = K (X-2)
That is, kx-y + 3-2k = 0
The distance from the center of the circle to the straight line is equal to the radius, that is: | 3K + 1 + 3-2k | / radical (k ^ 2 + 1) = 1
(k+4)^2=k^2+1
8k=-15
k=-15/8
The equation is y = - 15 / 8 (X-2) + 3
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