In the quadrilateral ABCD, ∠ a = 140 °, d = 80 °, if ∠ B = ∠ C, try to find out the degree of ∠ C 1) As shown in Figure 1, if ∠ B = ∠ C, try to find out the degree of ∠ C (2) As shown in Fig. 2, if the angle bisector be of ∠ ABC intersects DC at point E and be ‖ ad, try to find the degree of ∠ C; (3) As shown in Fig. 3, if the bisectors of angle ABC and BCD intersect at point E, try to find the degree of ∠ bec

In the quadrilateral ABCD, ∠ a = 140 °, d = 80 °, if ∠ B = ∠ C, try to find out the degree of ∠ C 1) As shown in Figure 1, if ∠ B = ∠ C, try to find out the degree of ∠ C (2) As shown in Fig. 2, if the angle bisector be of ∠ ABC intersects DC at point E and be ‖ ad, try to find the degree of ∠ C; (3) As shown in Fig. 3, if the bisectors of angle ABC and BCD intersect at point E, try to find the degree of ∠ bec

1：∠C＝70
2：∠C＝60
3：∠BEC＝110

In the circle inscribed quadrilateral ABCD, ∠ A: ∠ B: ∠ C = 2:3:4, find the degree of ∠ D In the circle inscribed quadrilateral ABCD, AC bisects BD vertically, ∠ BCD = 80 ° and calculates the degrees of the other three angles of the quadrilateral ABCD These are two questions

Is this a mistake? Is there a solution?

Given the triangle ABC, angle a = half angle B = one third angle c, try to find the degree of angle C

Angle a = half angle B
Angle B = 2 angle a
Angle a = one third angle c
Angle c = 3 angle a
Angle a + angle B + angle c = 180
Angle a + 2 angle a + 3 angle a = 6 angle a = 180
Angle a = 30
Angle c = 3 angle a = 3 * 30 = 90

In the triangle ABC, angle a = half angle B = 3 angle c, calculate the degree of angle a, B, C

Let ∠ a = x, so ∠ B = 2 ∠ a ∠ C = 3 ∠ a, and because ∠ a + ∠ B + ∠ C = 180 °, therefore, 6 ∠ a = 180 °∠ a = 30 °∠ B = 60 °∠ C = 90 °

The ratio of the inner angle degrees of a triangle is 1; 2; 3, and the degree of the largest angle is?

The degree of the largest angle is 180 * 3 / (1 + 2 + 3) = 90 degrees

A right triangle ABC, where b = 90 degrees, the degree of a is four times that of C

Angle a = 90 °× 4 / (4 + 1) = 72 °
Angle c = 90-72 = 18 degrees

In the right triangle ABC, if the angle c is 90 degrees and the ratio of angle a to angle B is 5 to 4, then the degree of angle a is There is another one: if the base length of an isosceles triangle is 12 and the length of the center line on the bottom edge is 8, then the distance from the center line of the bottom edge to the waist is equal to

1.∠A=90°×5/（5+4）=50°
2. ∵ the three lines of isosceles triangle are in one, that is, the height on the bottom edge, the center line on the bottom edge and the bisector of the top angle are a straight line
The center line divides the triangle into two congruent right triangles, and the two right sides of the right triangle are 8 and 12 / 2 = 6 respectively, and the hypotenuse is the waist of the isosceles triangle
According to Pythagorean theorem:
Waist 2 = (12 / 2) 2 + 8
♀ waist = 10
Let the distance from the center line of the bottom edge to the waist is equal to H. according to the area formula, the equation is as follows:
（1/2）*12*8=(1/2)*10*h*2
　∴h=4.8
That is: the distance from the center line of the bottom edge to the waist is equal to 4.8
ν select B

In the right triangle △ ABC, if

According to the meaning of the title
∠A+∠B=90
∠B-∠A=15
Subtracting two formulas
2∠A=75
∠ a = 37.5 degrees
This case is ∠ C = 90 degrees
When ∠ B = 90 degrees, ∠ a = ∠ B-15 = 75 degrees
Note: ∠ a cannot be 90 degrees

In RT △ ABC, ∠ C = 90 ° AB = 2Ac, the degree of ∠ A and ∠ B is obtained

As shown in the figure, ∵ C = 90 °, ab = 2Ac,
∴∠B=30°，
∴∠A=90°-∠B=90°-30°=60°．

In RT △ ABC, ∠ C = 90 ° AB = 2Ac, the degree of ∠ A and ∠ B is obtained

As shown in the figure, ∵ C = 90 °, ab = 2Ac,
∴∠B=30°，
∴∠A=90°-∠B=90°-30°=60°．