。The ratio of angle A angle B angle C in the acute quadrilateral ABCD is 4 3 2 B d 180 Find D

Set the angle dx, then there is:

x+(180-x)×(4+3+2)÷3=360;

x+(180-x)×3=360;

2x=180;

x=90;So the angle d90°

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The sum of the internal angles of the quadrilateral is 360°

b+∠d=180°

So a+ c=180°

a: b: c=4:3:2 i.e. a: c=4:2 b: c=3:2

a=180°*4/6=120°

c=180°*2 6=60° or x=180°-160°=60°b=60°*3 2=90°

d=180°-90°=90°

The right side is for me to ask for more, just ask for d, and write these on the left.

From the known, let the degree of angle a be 4x, then according to the proportional formula, the angle b=3x, the angle c=2x, because b+ d=180, so d=180-3x, according to the sum of the internal angles of the quadrilateral is equal to 360, that is, the sum of the angles a, b, c, d is 360, list an equation, solve x=30, substitution, the angle d=180-3x=180-3*30=90

For similar problems, you can set the unknowns like this and list an equation to solve.

Because b+ d=180° (known).

So b+ d= a+ c, b: d= a: c,3:n (representing the ratio of d in this quadrilateral) = 4:2 (quadrilateral inner angle and 360°).

i.e. a: b: c:

d=4：3：2：

3,b: d=3:3 and because b:

d=3:3 (verified), b+ d=180° (known), so d=180° 2=90°

The first level of junior high school, I believe you can understand, if so, please ask, I wish you progress in learning, satisfied and remember!

The right-angled ladder permeates the absolute shape, and the sum of the internal angles of the quadrilateral is equal to 360°.

Let the corners of the quadrilateral be 3x, 3x, 2x, and 4x respectively

3x+3x+2x+4x=360°；x = 30 ° a = 90 °, b = 90 °, c = 60 °, d = 120 ° a = b = 90 ° (can be proved parallel), c is not plex height equal to d so it is a right-angled trapezoid.

Summary. I'm glad to answer for you, it's a quadrilateral.

In the quadrilateral ABCD, if the angle A + angle B = 180 and the angle C + the angle D = 180, then the quadrilateral (so that it is not, not necessarily.

I'm glad to answer for you, it's a quadrilateral.

But it doesn't have to be a parallelogram.

Hope mine is helpful to you.

Let the angle b be x, then a is x 40, c is x, that is, x x 40 x 10 30 360, then 3x 60 360, x 100. then ab, cd, 140, 100, 90, 30, respectively.

What is called "adjacent outer corner"? Which angle that adds up to 180 degrees with the inside angle?

Procedure: The four inner angles of the quadrilateral are combined to form 360 degrees.

So angle a = 2 * 360 (2 + 3 + 5 + 8) = 40 degrees angle b = 3 * 360 (2 + 3 + 5 + 8) = 60 degrees angle c = 5 * 360 (2 + 3 + 5 + 8) = 100 degrees angle d = 8 * 360 (2 + 3 + 5 + 8) = 160 degrees outer angle ratio = (180-40) :(180-60) :(180-100):

1 is detailed enough.

Because the angle a: angle b: angle c: angle d = 2: 3: 5: 8 and because the angle a + angle b + angle c + angle d = 360°

So a=360°*2 18=40°

b=360°*3/18=60°

c=360°*5/18=100°

d=360°*8/18=160°

So the outer angle of a = 180°-40° = 140° and the outer angle of b = 180°-60° = 120°

The outer angle of c = 180°-100°=80°

The outer angle of d = 180°-160° = 20°

So the outer angle of a: the outer angle of b: the outer angle of c: the outer angle of d = 140°:120°:80°:20°=7:6:4:1

Set the angle a=2x then b=3x c=5x d=8x a+b+c+d=360 x=20, so a=40 b=60 c=100 d=160

The outer angle of a = 180-40 = 140 In the same way, the outer angle of b = 120 c = 80 d = 20 So the ratio is 7:6:4:1

Because the inner angle of the quadrilateral ABCD is 360°, the angle A: angle B: angle C:

Angular d=2:3:5:

8, so the angle a=40°, the angle b=60°, the angle c=100°, the angle d=160°, then the adjacent outer angles of angle a, angle b, angle c, and angle d are 140°, 120°, 80°, 20° respectively, and their ratio is 7:6:4:1.

The sum of the inner angles of the quadrilateral is 360 degrees.

So the angles a, b, c, and d are 90 degrees, 120 degrees, 90 degrees, and 60 degrees, respectively, so that the length of ab is x and the length of dc is y, then the area = 2*x 2+1*y 2

Angle a angle c is a right angle, Pythagorean theorem 2 2+x 2 =1 2+y 2 (2).

The angle d is 60 degrees, and the cosine theorem cos60 = (2 2+y 2-ac 2 ) (2*2*y) (3).

Angle b is 120 degrees cos120 = (x 2+1-ac 2) (2*1*x) (4).

Simplify(3)(4)2*x+y = 3+y2-x2;

y2-x 2 = 3 from (2);

So area = x+y2 = 3

There are two angles that are 90 degrees, one is 120 degrees and the other is 60 degrees.

The area is just settled, come by yourself.

1.According to the sum of the angles between the remaining two parallel lines is 180°, we can get a+ b=180°, b+ c=180°, c+ d=180°, d+ a=180°, and the diagonals of the parallelogram are equal

2.It is diamond-shaped. (From the knowledge of congruent triangles and the vertical fingers of the midpoint, it can be concluded that the quadrilateral is equal on all sides, and the quadrilateral with equal sides is a diamond.) ）

3.Add the condition of be=df. (Because be=df, ab=cd, abe= cdf, so abe cdf, so ae=cf.)

The same can be said for af=ce. And because the quadrilateral with equal sides is a parallelogram, a quadrilateral reed finger shape can be obtained, and AECF is a parallelogram. ）

4.Octagon. (n = 8 is solved according to the polygon inner angle and the formula 180°(n-3)=1080).

1 Blind or sorry 1 The topic is not very good, Qi Xiao understands...

2 parallelograms.

3ae = cf (the answer is not the same).

4 octagons.

Extend AB as an auxiliary line and cross DC at point H.

Let the angle a degree be 2x, then the angle d is 3x, and the angle b is 4xab The angle CBH is complementary to the angle ABC The angle CDE is complementary to the angle of CHE.

In the triangle CBH, the sum of the inner angles is 180°

So angle a + angle cbh + angle chb = 180

2x+(180°-3x)+(180-4x)=180° gives x=36°

So the angle a = 2x = 72°

Because: quadrilateral.

So: the sum of the four corners of ABCD = 360

Because: angle a: angle b: angle c: angle d=2:1:1:2 so: a = d = 120°, b = c = 60°

So: a+b=c+d=180°, side ab=cd so: side ad is parallel to side bc, i.e., this quadrilateral is isosceles trapezoidal.