How to find binary quadratic equations in MATLAB I have this result in my calculation syms x y; >> solve('x^2+y^2+30*y=175','x^2-90*x+y^2=-800') ans = x:[2x1 sym] y:[2x1 sym] It seems that this x and y are not the values I want. What's more, I can't understand them,

How to find binary quadratic equations in MATLAB I have this result in my calculation syms x y; >> solve('x^2+y^2+30*y=175','x^2-90*x+y^2=-800') ans = x:[2x1 sym] y:[2x1 sym] It seems that this x and y are not the values I want. What's more, I can't understand them,

Wrong use of solve function
It should be
solve('x^2+y^2+30*y=175','x^2-90*x+y^2=-800','x,y')
Add variables after, otherwise matlab does not know who to solve

How to use matlab to solve a system of binary quadratic equations?
The title is as follows: given that the coordinates of a, B and C are (0100) (0,0) (200,0), and the distance between an unknown point and ABC is 89100134, how to use matlab code to calculate the coordinates of the point?
I write a = solve ('x ^ 2 + (y-100) ^ 2-89 ^ 2 ','x ^ 2 + y ^ 2-10000', '(X-200) ^ 2 + y ^ 2-134 ^ 2');
But the prompt is wrong. How to write it?

>> [x,y]=solve('x^2+(y-100)^2-89^2','x^2+y^2-10000');>> x=double(x),y=double(y)x =   79.7022  -79.7022y =   ...

Matlab solution of binary quadratic equations!
(x+100)^2+y^2-1.9696xy-196.96y-90000=0
(x-100)^2+y^2-1.4142xy+141.42y-90000=0
use
syms x y;
solve('y^2+(x+100)^2-1.9696xy-196.96y-90000=0','(x-100)^2+y^2-1.4142xy+141.42y-90000=0','x','y')
How to prompt by pressing enter
Error using ==> solve>getEqns at 202' y^2+(x+100)^2-1.9696xy-196.96y-90000=0 ' is not a valid expression or equation.
Error in ==> solve at 67[eqns,vars] = getEqns(varargin{:});

It should be
1.9696xy middle multiplication sign: 1.9696 * x * y
196.96y 1.4142xy 141.42y, too

The angle between line and plane in space rectangular coordinate system
For example, in the regular triangular prism abc-a'b'c ', AA' = AB = 1, PQ are the points on the side edges BB 'and CC' respectively. The tangent makes the length AP + Po + QA 'of apqa' the shortest. Find the cosine value of the angle between AP and plane a'pq
The picture is a regular triangular prism, which is a rectangle composed of three squares. The shortest broken line apqa is the diagonal of the rectangle
Taking the midpoint of BC as the origin, BC as the Y axis and OA as the X axis, the space rectangular coordinate system o-xyz is established
I can only say so much. I don't know if I can draw the picture by myself
For the given answer, first find the cos value of the normal vector of PA and a'pq
Then the sin value of this angle is the cosine value. I don't know how to get it

The angle between the normal vector of the plane and the vector PA, the angle between the plane a 'PQ and PA, the sum of which is 90 degrees. This is a set of induced formulas

Optional 2-1 point to plane distance space coordinate system
Elective course 2-1, in the space coordinate system, what is the formula for finding the distance between a point and a plane? We need to use the normal vector of a plane
It is known that ABCD is a square, PD is perpendicular to ABCD, PD = ad = 2. Find the distance from point d to plane PAC

If only the distance is required, the volume can be used
The formula is a * b = | a | * | B | cos @, which is not the standard
The latter part "|b|cos @" is equivalent to the height (the formula here is really hard to play.)

How to calculate the distance between any two points on the earth's surface by longitude and latitude?
Hope the result is more accurate. It is mainly used to calculate the distance between two points from GPS coordinates

Let the radius of the earth be r, the center of the earth be 0, the spherical coordinates of two points a and B on the sphere be a (α 1, β 1), B (α 2, β 2), α 1, α 2 ∈ [- π, π], β 1, β 2 ∈ [- π / 2, π / 2], ab = R & # 8226; arccos [cos β 1cos β 2cos (α 1 - α 2) + sin β 1sin β 2] make use of Pythagorean theorem and sine theorem

If you know the longitude and latitude, orientation (direction angle) and distance of one point, how to calculate the longitude and latitude of another point?
30 | solution time: 2010-11-20 22:09 | questioner: zzb83
For example: I am at point a (113.00000, 22.00000), I walk 10km to the Northeast (45 degrees), what is the longitude and latitude of the destination? How to use the formula?
What are the methods or formulas in MapInfo or MapBasic?
If you know the circumference of the earth's equator (40000), then one degree will know how long it is. Due to different latitudes, you need to consider the angle between the earth and the equator. For example, 60 degrees is equal to half of the equator. The longitude is basically the same. If one circle is long, you can calculate how many degrees it is!
What's Da An, please? I haven't found it yet

First, change the distance into degrees, which is more accurate
The circumferences of the earth are almost equal, so it can be seen that walking 111 km is walking 1 degree
2 calculate the degree, n = 10 / 111
3 calculate the longitude, x = cos (45) * n
Latitude, y = sin (45) * n
4 longitude, x = 111 + X
Latitude, y = 22 + y

Find a formula in excel that can calculate the distance between two points according to the longitude and latitude of two points,
S==6371000*ACOS(COS(RADIANS(SUM(1*LEFT(C322,FIND("°",C322)-1),MID(C322,FIND("°",C322)+1,FIND("′",C322)-FIND("°",C322)-1)/60,RIGHT(LEFT(C322,LEN(C322)-1),LEN(C322)-FIND("′",C322)-1)/3600)-SUM(1*LEFT(A322,FIND("°",A322)-1),MID(A322,FIND("°",A322)+1,FIND("′",A322)-FIND("°",A322)-1)/60,RIGHT(LEFT(A322,LEN(A322)-1),LEN(A322)-FIND("′", A322)-1)/3600)))*COS(RADIANS(SUM(1*LEFT(D322,FIND("°",D322)-1),MID(D322,FIND("°",D322)+1,FIND("′",D322)-FIND("°",D322)-1)/60,RIGHT(LEFT(D322,LEN(D322)-1),LEN(D322)-FIND("′",D322)-1)/3600)-SUM(1*LEFT(B322,FIND("°",B322)-1),MID(B322,FIND("°",B322)+1,FIND("′",B322)-FIND("°",B322)-1)/60,RIGHT(LEFT(B322,LEN(B322)-1),LEN(B322)-FIND("′", B322) - 1) / 3600). This formula has errors. Take the following points for experiment
The first point: 118.9453 39.67206
The second point: 119.2804 39.58189
Using the above formula to calculate 38km, you can measure 30km on Google Earth. Who knows why? How to solve it

The formula for calculating the distance between any two points on the earth is as follows:
D＝R* arccos(siny1siny2+cosy1cosy2cos(x1-x2) )
Where R is the radius of the earth, with an average of 6370km
The longitude and latitude of point a are X1 and Y1, respectively. The east longitude is positive and the west longitude is negative
The longitude and latitude of point B are x2 and Y2, respectively. The north latitude is positive and the south latitude is negative
The distance between the two points is 30.4km, which is consistent with Google Earth
Note that latitude and longitude are angles. When calculating sin and COS, convert them into radians first

According to latitude and longitude, how to find the distance between any two points on the earth
Requiring precision or accuracy
Write a mathematical formula
I've written an accuracy report, which feels much worse. For some places, it's 10km worse
These five points are the only ones I have left
That's all I have

Suppose that the earth is a standard sphere with radius r, and that east longitude is positive, west longitude is negative, north latitude is positive and south latitude is negative,
Then the coordinates of a (x, y) can be expressed as (R * cosy * cosx, R * cosy * SiNx, R * siny)
B (a, b) can be expressed as (R * CoSb * cosa, R * CoSb * Sina, R * SINB)
So, the cosine of the angle that ab stretches to the center of the ball is
cosb*cosy*(cosa*cosx+sina*sinx)+sinb*siny=cosb*cosy*cos(a-x)+sinb*siny
So the spherical distance between two points AB is
R*{arccos[cosb*cosy*cos(a-x)+sinb*siny]}
Note: 1. X, y, a, B are all angles. The arccos given in the final result is in radian form
2. The so-called "east longitude is positive, west longitude is negative, north latitude is positive and south latitude is negative" is for the convenience of calculation
For example, if a certain point is 145 ° in Xijing and 36 ° in south latitude, then (- 145 ° and - 36 ° can be used for calculation
3. The ball method of the angle of AB to the center of the ball is actually to sum the angle K between two vectors
Use the formula * = | OA | * | ob | * cosk
You can get it

A formula for the distance between any two points on the earth, expressed in latitude and longitude (complete formula)

First of all, suppose that the earth is a standard sphere, its radius is r, ignoring the influence of terrain on distance. Let the longitude of point a be α 1 and the latitude be β 1; let the longitude of point B be α 2 and the latitude be β 2. At the same time, it is agreed that east longitude is positive and west longitude is negative; south latitude is 90 ° + geographical latitude, north latitude is 90 ° - geographical latitude, d = rarccos (c)