Urgent 50 Elementary School Math and Geometry Solving Questions 5

In order to make good use of this condition, we first make a 30-degree angle of the right triangle: do EM vertical CD in M, do CE perpendicular bisector CD in P, then the triangle EPM is a 30-degree angle of the right triangle, and EP=CP

If EN is perpendicular to BC and N, then EN=cm

If M is the midpoint of cd, then em=cm=half the length of the side of the square.

In this way, let em be a, then mp=root3 a, ep=cp=2a, so as to find cm, en length, square side length, and then find bn length, so that the tangent value of the angle ebc can be found, is en bn, and the result is equal to root 3 3, so we can know that the angle epc = 30 degrees.

Do CD bisector to hand over CD to P" changed to "do CE perpendicular bisector to hand over CD to P", I wrote it wrong before, and today I was idle and looked at my own findings that there was a mistake, and I changed it easily.

30 degrees. #include

#include

using namespace std;

int main(){

double a;

The angle is turned into a rotational radian.

a = ;Let the width and height of the square be 1

Half of it is a unit.

double ef = ;

printf("> ef length: %lf",ef);

Do the perpendicular line from E to BC as EF

double cf = ef * tan(a);

printf("> CF length = lf",cf);

double bf = 1-cf;

printf("> bf length = lf",bf);

double ebc = atan(ef/bf)*(180/;

printf("> EBC = D degree",(int)ebc);

return 0;

EF Length:

cf length =

bf length =

EBC = 30 degrees.

Solution: Take ac as the edge, make an equilateral triangle ace to the right, and connect be so ac=ae=ce

Angular cae = angular ace = 60 degrees.

Because the angle cab = angle CBA = 50 degrees.

So ca=cb

Because the angle cab + angle CBA + angle ACB = 180 degrees.

So the angle acb = 80 degrees.

Because the angle eab = angle cae - angle cab = 60-50 = 10 degrees.

Angular OAB = 10 degrees.

So the angle oab = angle ab = 10 degrees.

ce=cb, so the angle ceb=angular cbe

Because the angle ECB = angle ACB - angle ace = 80-60 = 20 degrees.

Angle ECB + Angle CEB + Angle CBE = 180 degrees.

So the angle cbe = 80 degrees.

Because the angle oba = angle CBA - angle obc = 50 degrees.

Angular obc = 20 degrees.

So the angle oba = 50-20 = 30 degrees.

Because the angle abe = angle cbe - angle oba = 80-50 = 30 degrees.

So the angle oba = angle abe = 30 degrees.

Because ab=ab

So the triangle oba and the triangle eba congruence (asa) so oa = ae

So ae=ac

So the angle OCA = the angle AOC

Because the angle OCA + angle AOC + angle OAC = 180 degrees.

Angle OAC = Angle CAB - Angle OAB = 50-10 = 40 degrees.

So the angle oca = 70 degrees.

It turns out that the circumference is not possible, which is obviously only possible with the known AE length.

I offer a way to solve the problem.

Tell the teacher that this year's three good students are you!

cm is not perpendicular to ab to m. If so, the red line part is incorrect. It should be the square of the cm and not the square of the ch.

I don't understand why, I did it.

Solution: Extend the ED, and AB is handed over to G

Angle dec = 30 degrees, angle bac = 60 degrees.

Angular age = 90 degrees, i.e., eg perpendicular ab

BC is perpendicular to AE, and BC intersects EG at D

d is the vertical center of the triangle ABE.

and af over d and be over f

AF Vertical BE

CB=CB1, the triangle CBB1 is an isosceles triangle, the angle ABC=90-A=angle A1B1C1=angle CBB1, so the angle BCB1=2A=angle ACA1=X

Uh......Don't take the class seriously.

The title won't be.

I'm in my third year of junior high school, and I have to think about the topic for myself.

Otherwise, taking the high school entrance exam you are annoyed.

Solution: In ABC, rotate ABC x° to A,B,C BC=B with C as the center'c ∠a=∠a'=a ∠acb=90°∴∠b'=∠cbb'

a'b'c, b'=90°-a

bb'C, BCB'=180°-2（90°-a）=2a

Because it is rotating, cb=cb', so b'＝∠cbb'＝90°－∠a＝90°－a；So x= bcb'=180°-∠b'-∠cbb'=2a.

It should also be stated that point E is on BC, otherwise it cannot be proven. When the point E is on BC, the proof method is as follows:

Let AD and BC intersect M (if D is on BC, M and D coincide).

bac＝∠adc＝90°，∴abc＝∠cad。The co-angle of the same is mbh abc 2, mah cad 2, mbh mah, a, b, m, h are round, ahb adb 90°.

From AHB 90°, ABH EAH, we get: Gh is the perpendicular bisector of AE, AG EG, AEG DAE, and DAE CAE, AEG CAE, EG AC.

I don't want to draw a picture.