It is known that the area of triangle ABC is s, the acute angle between plane ABC and plane a is Q, the projection of triangle ABC on plane a is triangle a'b'c ', and the area of triangle a'b'c is s'. This paper proves that s' = s * cosq (please tell me the proof idea,

It is known that the area of triangle ABC is s, the acute angle between plane ABC and plane a is Q, the projection of triangle ABC on plane a is triangle a'b'c ', and the area of triangle a'b'c is s'. This paper proves that s' = s * cosq (please tell me the proof idea,

1. We can know s' = s * cosq when we transform a solid into a plane
2. S = 0.5A * H (a is the bottom, h is the high)
3. Make the projection of height. The projection of height is h * cosq
4. The bottom is unchanged, so s' = 0.5A * h * cosq
I suggest you draw a picture and understand it

PA is the tangent of circle O, a is the tangent point, Po intersects circle O and point B, PA = 8, OB = 6, then Tan angle apo

Connect Ao
⊥ PA is the tangent of circle O, and a is the tangent point
∴AO⊥AP
∵OB=OA=6,AP=8
∴tan∠APO=6/8=3/4
I understand it by myself. I'm not sure

If a tangent is drawn from a point on the straight line y = x + 1 to the circle (x-3) 2 + y2 = 1, the minimum length of the tangent is ()
A. 1B. 22C. 7D. 3

The minimum value of tangent length is obtained when the distance between the point on the straight line y = x + 1 and the center of the circle is the smallest, the distance between the center of the circle (3,0) and the straight line is d = | 3 − 0 + 1 − 2 = 22, the radius of the circle is 1, so the minimum value of tangent length is D2 − R2 = 8 − 1 = 7, so select C

What is the minimum value of the tangent length when the tangent is drawn from the point on the straight line y = x + 1 to the circle (x-3) ^ 2 + (y + 2) ^ 2 = 1?
Is it root number 17 or root number 19? The calculation result is controversial. If the answer is worked out, let the rookie know

It should be the root 17, the distance from the center of the circle to the straight line is 3, the root 2,
The radius is 1, the square of the tangent length + the square of the radius = the square of the distance from the point on the line to the center of the circle,
If the tangent length is the minimum, then the distance from the point to the center of the circle must be the minimum, so it must be the distance from the center of the circle to the straight line

If a tangent is drawn from a point on the line y = x + 1 to a circle (x-3) + y = 1, then the minimum length of the tangent is?

Since the distance d from the center of a circle (3,0) to the straight line y = x + 1 is equal to 2 √ 2 > R = 1, if the point on the line leads to the tangent of a circle and the tangent length is the smallest, it must be the point on the line with the smallest distance to the center of a circle

If a tangent line is drawn from a point on the straight line y = x + 1 to the circle (x-3) ^ + y ^ = 1, then the minimum value of the tangent line is
If a tangent line is drawn from a point on the straight line y = x + 1 to the circle (x-3) ^ + y ^ = 1, then the minimum value of the tangent line is?

Let the point on the line y = x + 1 be P
Tangent point a, center C (3,0)
Then Ca ⊥ pa
From Pythagorean theorem
Tangent length PA = √ (PC & # 178; - Ca & # 178;) = √ (PC & # 178; - 1)
So when PC is the smallest, PA is the smallest
The minimum value of PC is the distance from C to the line y = x + 1
So the minimum value of PC = | 3-0 + 1 | / √ 2 = 2 √ 2
So the minimum value of tangent length PA is √ [(2 √ 2) & # 178; - 1] = √ 7

If a tangent is drawn from a point on the straight line y = x + 1 to the circle (x + 3) ^ 2 + y ^ 2 = 1, the minimum length of the tangent is________ (process)

Let the center of the circle be C, the tangent point be B, and the point on the line be a. then ab ⊥ BC has ab & sup2; = AC & sup2; - BC & sup2; BC has a fixed radius. To make AB shortest, AC must be shortest, and AC represents the distance between point C and the point on the line

If a tangent is drawn from a point on the straight line y = x + 1 to the circle (x-3) 2 + y2 = 1, the minimum length of the tangent is ()
A. 1B. 22C. 7D. 3

The minimum value of tangent length is obtained when the distance between the point on the straight line y = x + 1 and the center of the circle is the smallest, the distance between the center of the circle (3,0) and the straight line is d = | 3 − 0 + 1 − 2 = 22, the radius of the circle is 1, so the minimum value of tangent length is D2 − R2 = 8 − 1 = 7, so select C

Given the point m (3,1), the line ax-y + 4 = 0 and the circle (x-1) ^ 2 + (Y-2) ^ 2 = 4, find the tangent equation of the circle passing through the point M

If the tangent slope does not exist, then the tangent is x = 3. If the tangent slope exists, set K, then the tangent is Y-1 = K (x-3), that is, kx-y + 1-3k = 0. Then the distance from the center of the circle to the tangent is the radius, so | K-2 + 1-3k | / √ (k ^ 2 + 1) = 2, so | 2K + 1 | = 2 √ (k ^ 2 + 1) the solution is k = 3 / 4, so the tangent is 3x-4y-5 = 0. In conclusion, there are two tangents

From the point m (2,3) to the circle (x-1) ^ 2 + (Y-1) ^ 2 = 1, the tangent equation is obtained

Center (1,1) of circle (x-1) ^ 2 + (Y-1) ^ 2 = 1, radius 1
The abscissa of point m (2,3) is 2, and the abscissa of the rightmost end of the circle is 1 + 1 = 2
The point m (2,3) is exactly the same as the abscissa of the rightmost end of the circle, so a vertical line to the x-axis through the point m (2,3) must be tangent to the circle
One of the tangents of a circle is X-2 = 0
Let the tangent slope of another circle passing through point m (2,3) be K, then the tangent equation (Y-3) / (X-2) = k, that is:
kx-y-2k+3=0
If a line is tangent to a circle, the distance from the center of the circle to the line is equal to the radius
|K-1-2k + 3 | / radical (k ^ 2 + 1) = 1
|2-k | / radical (k ^ 2 + 1) = 1
(2-k)^2 = k^2+1
k = 3/4
3 / 4x-y-2 * 3 / 4 + 3 = 0, that is, 3x-4y + 3 = 0
To sum up, the two tangent equations of a circle are as follows:
x-2=0
3x-4y+3=0