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In △ ABC, angle c = 90 °, angle cab = 2 times angle B, ad bisecting angle cab, AC = 6 times root sign 3?
According to the meaning of the title, angle cab = 60 degrees, angle CBA = 30 degrees; AC = 6 (radical 3), BC = 18, AC = 12 (radical 3)
Ad bisector angle cab, BD: DC = AB: AC = 2
BD=2DC,BD+DC=18
DC=6.
As shown in the figure, in the semicircle AOB, ad = DC, ∠ cab = 30 °, AC = 23, the length of ad is calculated
∵ AB is the diameter, ∵ ACB = 90 °, ∵ cab = 30 °, ∵ ABC = 60 °, ∵ arc BC degree = 12 arc AC degree; ∵ ad = DC, ∵ arc ad degree = arc DC degree = 12 arc AC degree, ∵ arc BC degree = arc ad degree; ∵ BC = ad
AB is the diameter of semicircle o, C is the point on semicircle o which is different from a and B, CD ⊥ AB, perpendicular foot is D, ad = 2, CB = 4 * radical 3, then CD = -?
Let AC = y, BD = x, then there is
Y^2+48=(2+X)^2
Y^2-4=48-X^2
Get (x + 1) ^ = 49 (negative value rounding off)
x=6
CD=√(48-36）=2√3
The root sign can be replaced by a check in the mathematical symbol
As shown in the figure, in the RT triangle ABC, the angle B is equal to 90 degrees, and D is a point on ab. take the semicircle o of BD diameter tangent to AC and the point E, BD = BC = 6, to find the length of the hypotenuse AC
When OE is connected, then OE ⊥ AC, ⊙ AEO = ∠ B = 90 °, a = ∠ a, ∵ Δ AEO ∽ Δ ABC, ∵ OE / BC = AE / AB, 3 / 6 = AE / AB, let AE = x, then AB = 2x, AC = 6 + X, in RT Δ ABC, AC ^ 2 = AB ^ 2 + BC ^ 2, (6 + x) ^ 2 = (2x
As shown in the figure, ab = 10, the center O of the circle is the midpoint of AB, AE and BF are tangent lines, satisfying AE = BF. Take the moving point G on the arc EF to cross g and make the tangent line of the circle AE and BF
When point G moves, let ad = y, BC = x, then the relation between Y and X is y=
What about the picture?
As shown in the figure, AC ⊥ AE, BD ⊥ BF, ∠ 1 = 35 ° and ∠ 2 = 35 °, verification: AE ∥ BF
It is proved that: ∵ AC ⊥ AE, BD ⊥ BF, ∵ 1 + ⊥ 3 = ∵ 2 + ⊥ 4 = 90 °, ∵ 1 = 35 °, ∵ 2 = 35 °, ∵ 3 = ∵ 4, ∵ AE ∥ BF
If two numbers are randomly taken out in the interval (0,1), the probability that the sum of the two numbers is less than 6 / 5 is?
In the plane rectangular coordinate system, make a square with (0,0) (0,1) (1,1) (1,0) as the vertex. Then all values may be in the square. The sum of the two numbers is less than 6 / 5, which is the part of the square below the straight line x + y = 6 / 5 (to the left). Through the image, we can know that the area is 1-1 / 2 * (4 / 5) ^ 2 = 17 / 25, and the square area is 1. So the probability is (17 / 25) / 1 = 17 / 25
X belongs to [0,2 beats], find SiNx = cosx, SiNx > cosx, SiNx
sinx=cosx
tanx=1
X = π / 4 or x = 5 π / 4
sinx>cosx
1.cosx>0
sinx/cosx>1
tanx>1
x∈(π/4,π/2]
2.cosx
Take any number in the interval [0,3], which is the square of inequality X - ax + 2 ＜ 0, and the probability of a solution is 1 | / 3, then a =?
When x is 0, the inequality is obviously not tenable, and then 1,2,3 are brought in. There is only one inequality tenable. According to this condition, we can find a, but what a gets is a range
In the interval [0,10], a number x is randomly selected to make the inequality x2-2x
x²-2x
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