It is known that: as shown in the figure, in △ ABC, D is a point on the Ba extension line, AE bisects ∠ DAC, and AE ‖ BC. It is proved that ∠ B = ∠ C

It is known that: as shown in the figure, in △ ABC, D is a point on the Ba extension line, AE bisects ∠ DAC, and AE ‖ BC. It is proved that ∠ B = ∠ C

Proof: ∵ AE ∥ BC,
∴∠DAE=∠B，∠EAC=∠C，
∵ AE bisection ∵ DAC,
That is ∠ DAE = ∠ EAC,
∴∠B=∠C．

As shown in the figure, in △ ABC, ab = AC, D is a point on Ba extension line, AE bisects ∠ DAC, so AE ‖ BC? Tell me your reasons

AE∥BC．
∵AB=AC，
∴∠B=∠C，
∵ AE bisection ∵ DAC,
∴∠DAE=∠CAE，
And ? DAE +  CAE =  B + ∠ C,
ν 2 ∠ DAE = 2 ∠ B, i.e. ∠ DAE = ∠ B,
∴AE∥BC．

As shown in the figure, AE is the bisector of the outer angle ∠ DAC of △ ABC, and AE ‖ BC indicates that △ ABC is an isosceles triangle

Because of the fact that ‖ AE is equal to the angle
∠ EAC = ∠ ACB (internal staggered angle)
And ∠ ead = ∠ eac
So ∠ ABC = = ∠ ACB
Delta ABC is an isosceles triangle

If the radius of circumscribed circle and inscribed circle of a right triangle are 5 and 2 respectively, the tangent value of the smaller acute angle in the right triangle is

If the tangent value is 3 / 4, let the right triangle ABC, the angle c be the right angle, and the shorter side AC. because, the circumscribed circle radius is 5, that is, the oblique side AB is 10, let the center of the inscribed circle be I, and make a vertical line to AC and BC through I, and the perpendicular foot is e and F. because the angle c is a right angle and the radius of the inscribed circle is 2, ie = if = EC = FC = 2

If a right triangle has two sides of 6 and 8, the tangent of the smaller acute angle is?

Two cases
When 6 and 8 are right angles
The tangent of the minimum angle is 3 / 4
When 8 is a hypotenuse, the other right angle is 2 √ 7
The tangent of the minimum angle is √ 7 / 3

If a right triangle has two sides 7 and 4, what is the tangent of the smaller acute angle?

If 7 is a right angle side
The tangent of the smaller acute angle is 4 / 7
If 7 is bevel
Then the right angle side is √ (7? - 4?) = √ 33
So the tangent of the smaller acute angle is = 4 / √ 33 = 4 √ 33 / 33

If a right triangle has two sides of length 4 and 7, the tangent of the smaller acute angle is

If a right triangle has two sides of length 4 and 7, the tangent of the smaller acute angle is 4 / 7 or 4 √ 33 / 33

In a right triangle, the hypotenuse is four times of the height on the hypotenuse

Two smaller right triangles divided by the height on the hypotenuse are similar triangles
Let the height divide the hypotenuse into x and y
Then x: 1 = 1: y
And X + y = 4
It can be calculated that x = 2 + change 3, y = 2 - change 3
Or x = 2-change 3, y = 2 + change 3
Take the smaller part as X
Then the tangent of the larger acute angle is 1 / (2-change 3) = 2 + change 3
It's very clear to draw a sketch by yourself

How can the polar coordinate form of complex number a = 10 ∠ - 60 degree be transformed into algebraic form?

a=| F |cosq,b=| F |sinq
A=a+jb
A=10*cos(-60)+j10*sin(-60)

How to calculate this? Is it to change the polar coordinates into complex numbers or to change the following complex numbers into polar coordinates?

All right, but it is more convenient to use polar coordinates when calculating electrical quantities by phasor method, because j31.4 is written in polar coordinate form, that is, the modulus value is 31.4, the argument angle is 90 degrees, and the final result is module division and argument subtraction