As shown in the figure, ∠ 1 and ∠ 2, ∠ 3 and ∠ 4 are formed by which straight line? What are the angles of their respective positions

As shown in the figure, ∠ 1 and ∠ 2, ∠ 3 and ∠ 4 are formed by which straight line? What are the angles of their respective positions

(1) CD, AB by BD, stagger angle ad, BC by BD, stagger angle (2) AB, CD by BC, ipsilateral angle ad, BC by AB, isoangle

As shown in the figure, which two lines are formed by which straight line and which angles are they?

Left:
Two lines, AB and CD, are intersected by the straight line BD
3 and 4 are two lines AD and BC, which are intersected by straight line BD
Right:
1 and 2 are the internal angles of AB and CD cut by BC
3 and 4 are the apposition angles of two lines AD and BC cut by AE

In a triangle, there is an angle of 30 degrees, and its opposite side is equal to half of one of the sides
Please use the knowledge of circle and explain the reason

It is known that △ ABC, ∠ a = 30 degrees, AC = 2BC
Do △ ABC circumscribed circle O, do CO intersecting circle at D, we can know ∠ CDB = ∠ a = 30 degrees, ∠ CBD = 90, BC = R,
That is, AC = 2R is the diameter of the circle and △ ABC is a right triangle

In a triangle, there is an angle of 30 degrees, and its opposite side is equal to half of one of the sides. Can this show that the triangle is a right triangle

It shows that this triangle is a right triangle
But this is not a theorem and can not be applied directly

It is proved that an acute angle of a triangle is 30 ° and its opposite side is equal to half of the longest side. It is a right triangle with 30 ° angle
. please
. just today...

Let C be the largest edge, C = 2A. Let ∠ B = x, then ∠ C = 150 ° - X. then, according to the sine theorem, a / Sina = C / sinc. Since C = 2A, a = 30 °, so Sina = 1 / 2. Substituting these terms into the above formula, sin (150 ° - x) = 1, so 150 ° - x = 90 ° and x = 60 °

If the opposite side of 30 degrees in a triangle is half of the other side, then the triangle must be a right triangle

Let 30 degrees be X
Then the other side is 2x and its diagonal is a
According to the sine theorem
x/sin30=2x/sinA
sinA=1
A=90
So it must be a RT triangle

In a triangle, if an angle is 30 ° and the opposite side of the 30 ° angle is half of the other side, is the triangle a right triangle?
How can I prove it!

It can be judged
If we don't use trigonometric function to prove it, we can introduce circle to prove it
Suppose BAC = 30 ° and take point C as the center, make a circle with radius of AC / 2,
It can be proved that AB is the tangent of the circle (that is, AB is perpendicular to CB)
In other words, right triangles hold

The side corresponding to 30 degrees is half of the hypotenuse. Is this triangle a right triangle? Yes, how to prove that?

According to the sine theorem, if the length of the side opposite the 30 ° angle is a, then the hypotenuse is 2a, a / Sina = B / SINB. Then the sine value of the angle corresponding to the hypotenuse is 1, that is, the angle is a right angle, so

Can one side of the triangle be equal to half of the other side to determine whether it is a right triangle and determine the angle of 30 degrees?

This is obviously a wrong idea. We fix the long side and rotate the short side. There are infinite triangles

If a square equals 7 triangles, and a square times a triangle equals 63, then square = {=

Square = 21
Triangle = 3