In the quadrilateral, B D 90 C 30 BC 10 3 AD 4 3 find the AB and CD lengths

If you pass point B to do BE cd, to E, and to point A do AF to E, then BE=BC*sin30°=5 3 CE=BC*COS30°=15 and quadrilateral, ADEF is rectangular, EF=AD=4 3BF=BE-EF= 3

In the quadrilateral ABCD, B= D=90° C=30° Bad=150

baf=150°-90°=60°

ab=bf/sin60°=2 de=af=√（ab²-bf²）=1cd=ce+ed=1+15=16

Solution: If BA and CD are extended to E, then the triangle EBC and the triangle EDA are right-angled triangles.

Angle c = 30°, so be=bctan30°=10 1 3ae=ad sin dea=4 3 3 2=8, so ab=be-ae = ten-thirds of the root number three minus 8ce=bc divided by cos30°=twenty-thirds of the root number three.

ed=ae cos∠dea=4

So cd=ce-ed=20 times the root number 3 minus 4, I don't know if you understand?

By intersecting the points of Cd and F on the parallel lines of B as AD, then Bfc=90°, C=30°, then Bf=BC 2=5 3, ABFD is a right-angled trapezoid, ABF=30°, then AB=2, CD=CF+DF=(3 2)*10 3+AB 2=15+16.

To take full advantage of the relationship in the 30° RT, click **.

Do the auxiliary line: take a little E on BC so that Ed is perpendicular to AD, take a point F on AB so that EF is perpendicular to AB So, the angle ADE=90 degrees, the angle of the history of the oak AFE=90 degrees, so the paralimb quadrilateral AFED is a rectangle, AF=DE, AD=EF=1 quadrilateral ABCD, A= C=90°, B=60°, so the angle D is equal to 120 degrees in the triangle BEF, the angle B=6....

No solution, or few conditions.

First, make the length of the AD line segment equal to 4, and make two straight lines AM and DN through A and D respectively, so that the angle A = 30 degrees and the angle D = 120 degrees.

Then there are an infinite number of line segments BC in these two straight lines, so that BC=1 so that the position of point C is uncertain, then the length of CD is uncertain.

Through point C to do CE parallel to AB and intersection of AD and point E, from the meaning of the question, ECD=45 degrees, and because D=90 degrees, so cd=de=1, so AE=2

Then do EF perpendicular to AB at point F, because a=45 degrees, so EF=AF=BC=root number 2

To do the auxiliary line: take a little E on the BC so that the ED is perpendicular to AD, and take a little F on the AB so that the EF is perpendicular to the AB

So, the angle ADE=90 degrees, and the angle AFE=90 degrees, so the quadrilateral AFED is rectangular.

It is learned that af=de, ad=ef=1

In the quadrilateral ABCD, A= C=90°, B=60°, so the angle D is equal to 120 degrees.

In a triangular BEF, the angle b = 60 degrees and the angle bfe = 90 degrees, so the angle feb = 30 degrees.

Knowing one side of the triangle, solving the triangle, you can find bf, be

In the triangle CDE, the angle EDC = 120 degrees - 90 degrees = 30 degrees, so the angle dec = 60 degrees.

ec=bc-be, after obtaining ec, one side of the triangle is known, the triangle is solved, and ed, cd is obtained

So, ab=af+bf=ed+bf

Count with the root number, I won't write it for you if I don't write it, calculate it slowly and carefully, don't make a mistake.

10 times the root number 3 12 = x (self-counting), then ad=2x (the right angle of a right triangle with an acute angle of 30 degrees is equal to half of the hypotenuse).

Hello you first draw a diagram for this question, note that the two angles of b d are diagonal, mark the four angles of a, b, c d, and you draw the two lines of bc ad along the direction of b to c, a to d, and the intersection is set to m point, a=60°, b=90°, so m=30°, and then the nuclear shed ab=2, so you can find am=4, bm=2 3, and cdm=90°, cd=1, so cm=2, md= 3, so ad=4- 3, bc=2 3-2

ad=16-4*(root number 3).

bc = 8 * (root number 3) -8

Because you can't upload **, let's talk about the method briefly.

Make parallel lines of AD from point C and cross AB at point E, and make parallel lines of CD from point E and cross AD at point F.

A CDFE is a rectangle.

Angle EAD = 60 degrees, angle BCE = 30 degrees.

Let ae=x, eb=8-x

Then there is af=x 2

fd=ec=2*be=2*（8-x）

ad=af+fd=x/2+2*（8-x）

x=ae=ef*2 root number3=4*2 root number3=8 root number3 substitution yields ad=16-3x 2=16-4*root number3 in the same way: bc=root number3*be=root number3*(8-8 root number3)=8*(root number3)-8

In fact, it is very simple, it is a problem to find the length of the side in a right-angled trapezoid, and you can use the Pythagorean theorem.

You need to make an auxiliary line to cross the point d to ab to make an auxiliary line to intersect ab to m, according to ab=8, cd=4 can get am=4, and a 60° b= d=90°, you can get adm 30°, and then you can use the pythagorean theorem to calculate the value of the side ad, am bc