It is known that in △ ABC, ∠ BAC = 120 °, ad bisects ∠ BAC, ab = 5, AC = 3, and finds the length of AD

It is known that in △ ABC, ∠ BAC = 120 °, ad bisects ∠ BAC, ab = 5, AC = 3, and finds the length of AD

Let ad = x, ∵ ABC ∽ EDC, ∵ deab = ECAC, that is, X5 = 3 − X3, ∵ x = 158, ∵ ad = 158
Given the inverse scale function y = - 12 / x, find the value range of the independent variable x when y is less than or equal to 4 / 3
y < 3/4
-12 / x < 3 / 4 (x is negative)
-12 > 3x/4
-48 > 3x
x < -16
As shown in the figure, in the triangle ABC, ∠ ACB = 90 °, the angle bisector of ∠ CAD intersects the extension line of BC at E. if angle B = 50 °, calculate the degree of angle AEB
The picture does not come out, is the junior one synchronization exercise and the appraisal topic
If ∠ CAD is ∠ cab, the bisector should not be handed over to the extension line of BC, it should be handed over to BC directly. So there is a problem with the title
If the title is changed to "in the triangle ABC, the angle bisector of ∠ ACB = 90 °, the angle bisector of ∠ cab intersects BC at e, if the angle B = 50 ° find the degree of angle AEB", then
∵∠ACB=90°,∠B=50°
∴∠CAB=40°
∵ AE bisection ∠ cab
∴∠EAB=20°
∴∠AEB=180°-∠B-∠EAB=110°
What about the picture
The CAD should be cab
∠AEB=∠ACB+∠CAE=90°+∠CAB/2=90°+(90°-∠CBA)/2=90°+(90°-50°)/2=110°
Should CAD be cab?
∠AEB=180-∠B-∠EAB=180-∠B-1/2∠A=110°
I don't think the title is wrong! I did it, too.
If the point a (m, - 2) is on the inverse scale function y = 4 / X ', then when the function value is greater than or equal to - 2, the value range of the independent variable x is
Why is the answer that x is greater than 0 'and if it is greater than 0, it can be greater than - 2' but not equal to - 2?
According to you, that X has to be taken as - 2, because only one of the numbers where x ≤ - 2 is - 2, then X
As shown in the figure, △ ABC, the bisector BD of ∠ B and the bisector ce of the outer angle of ∠ C intersect at point P. verification: the distance from point P to the straight line of AB, BC and Ca on three sides is equal
It is proved that: as shown in the figure, the vertical line passing through point P is the straight line of AB, BC and Ca on three sides, and the vertical feet are Q, m and N respectively. Then the vertical line segment PQ, PM and PN is the distance from point P to the straight line of AB, BC and Ca on three sides. ∵ P is a point on the bisector BD of ∠ ABC, ∵ PM = PQ. ∵ P is a point on the bisector ce of ∠ ACM, ∵ PM = PN. ∵ PQ = PM = PN. ∵ P is the distance from point P to the straight line of AB, BC and Ca on three sides Equal
If point a (m, - 2) is on the image of inverse scale function y = 1 / x, then when function value y is greater than or equal to - 2, the value range of independent variable x is?
In y = 1 / X
If x = - 2, then when x0, Y > 0, Y > = - 2
So x ≤ m, x > 0
-2*m=1，m=-1/2
K = 1 > 0, y decreases with the increase of X in each quadrant, but Y > 0 in the first quadrant
When x ≤ - 1 / 2 or x > 0, Y > - 2
First find m, M = - 1 / 2, X is less than or equal to - 1 / 2 or X is greater than or equal to 0
As shown in the figure, in △ ABC, the bisector BD of ∠ B intersects the opposite extension of the bisector ce of the outer angle at point D. if angle a = 27 degrees, then angle d =?
According to the meaning of the title, in △ ABC, the bisector BD of ∠ B intersects the opposite extension of the bisector ce of the outer corner at point D, if ∠ a = 27 degrees
∵∠ D + 1 / 2 ∠ B = 1 / 2 external angle, i.e. 2 ∠ D + B = external angle
And the outer angle = ∠ B + ∠ a = ∠ B + 27 degree = 2 ∠ D + ∠ B, ∠ d = 1 / 2 ∠ a = 13.5 degree
Fourteen
As shown in the figure, in △ ABC, the bisector BD of ∠ B intersects the opposite extension of the bisector ce of the outer angle at point D. if angle a = 27 degrees, then angle d =?
It is known that y is the inverse proportion function of X. when x is equal to 2, y is equal to 0.5
The values of y when x is equal to 3 and X is equal to minus one third are obtained respectively
The formula is: y = 1 / x, x = x ≠ 0, x = 3, y = 1 / 3, x = - 1 / 3, y = - 3
It is known that CE is the angular bisector of △ ABC outer angle ∠ ACD, and CE intersects BA in E
It is proved that ∵ EC is the bisector of ∵ ACD, ∵ ace = ∵ ECD, ∵ ECD = ∵ B + ∵ e, ∵ ECD ∵ B, and ∵ BAC = ∵ e + ∵ ECA,