Math problems, to verify the angle, ask for help

EF cd angle EFB = angle DCB, so angle GDC = angle DCB, so dg bc, according to two parallel isotope angles equal so angle AGD = angle ACB

ef⊥ab，cd⊥ab

ef||cd (two straight lines perpendicular to the same line parallel to each other) efb = bcd (two straight lines parallel and equal at the same angle) efb = gdc

BCD = GDC (Equivalent Substitution).

dg||BC (equal internal wrong angles, two straight lines parallel).

AGD = ACB (two straight lines parallel with equal isotope angles).

Let the height of the large triangle be h, and the height of the small triangular shape be h. The bottom edge is divided into two parts by the perpendicular line, and the length of the left side is changed to A, and the length of the right side is B. The angle to be sought is .

Then there is a nuclear antan:

h a = tan10°, then h = a * tan10°h a = tan20°, then h = a * tan20°b h = tan40°, then b = h * tan40° = a * tan20° * tan40°

And because: tan = h b = a * tan10° (a * tan20° * tan40°).

So: tan = tan10° (tan20°tan40°) to solve this equation to get 30°

If the outer picture is a square, each corner of the lower square is 90 degrees, and the 45-degree angle of the base angle is exactly half of the 90-degree angle. Fold the vertices of the 45 degree angle of the square and the triangle in half, then the outside of the triangle and the inner part of the square can just cover the triangle, so the angle is equal to the unknown angle of the triangle: that is:

The angle sought is 72 degrees. If it's not a square, it seems like the conditions aren't enough. (The yellow and blue parts are folded over, which coincides with the triangle, because the square is square, so the two sides of the fold are equal, and the two right angles ensure that the two sides are in a line.) ）

Extend the lower left or upper right, take the equal line segment to prove the congruence, and the answer should be 72°

Triangle outer corner = sum of 2 inconsistencies not adjacent.

If it's inside a square, then it's an acute angle.

Two triangles of different colors are congruent Then the acute angles of these two triangles are equal 90-45=45 45 2= Then this angle is equal to .

The angle sought is 68 degrees, the upper one of the 45 degree angle is 22 degrees, the upper angle of the lower right triangle is 40, the lower angle is 50, the intersection of the blue triangle below is 67, and the upper angle is 23

This is a mathematical theorem that the number of angles outside a triangle is equal to the sum of the two inner angles that are not adjacent to it, so the answer is 45 + 63 = 108

The 3 corners of the triangle add up to 180 degrees. The two corners in a straight line are also 180. So it's 45 + 63 = 108 degrees.

45° below the angle is 5° and 40° above

63° is 85° on the left and 32° on the right.

bod=64

Because the ob and od are equally divided into each other, no matter how much aoe is, bod= lift the nucleus aoc 2+ coe 2= aoe 2

sin(θ-20º)=1/2=sin30º=sin150

Look at the angle in which the sine is equal to 1 2 in -20 340.

<> "is a theorem, the three angular bisectors of a triangle intersect at one point, and the angle is added and subtracted to obtain a regular triangle, and the calculation is completed, it is too simple, there should be no problem with the above solution!" It should stand up to automatic inspections at all times! But why was it reported in such a short period of time?

The ** that answered the question was just posted for an hour.,I received an apology short message that resumed the display.,And the ** that was posted before was only an auxiliary line but not answered** was hung for a day.,Who is so timely? I can only guess that it is the subject! Because only the subject can receive the message of the answer to the question in time!

Then this subject is too bad.

Lack of known conditions to make an answer.

Is AD parallel to BC?

Parallel then the angle sought is 100 degrees.