sin(x+pai/6)=1/4,sinx=?

sin(x+pai/6)=1/4,sinx=?

Cos / Pax = 6
Tan (x + Pai / 6) = plus or minus 1 / radical 15 = (TaNx + 1 / radical 3) / (1-tanx / radical 3)
The solution is TaNx = so and so
(tanx)^2 +1=(secx)^2=1/(cosx)^2
The solution is cosx = XX
So we get SiNx
I didn't figure it out
Let a and B be functions y = K / X. let a and B be any two points symmetrical about the origin o on the image. Let AC be parallel to the Y-axis and BC be parallel to the x-axis. Let a and B be known that the area of △ ABC is 4. Let K be found
Given that a and B are functions y = K / x, any two points on the image which are symmetric about the origin o, AC is parallel to the Y axis, BC is parallel to the X axis, given that the area of △ ABC is 4, find the value of K
Let a (x0, Y0), then B (- x0, - Y0),
y0=k/x0.k=x0*y0
The equation of AC is x = x0, and the equation of BC is y = - Y0
C(x0,-y0)
|AC|=|y0|,|BC|=|x0|
S△ABC=1/2|x0|*|y0|=4
x0*y0=±8
That is: k = ± 8
As shown in the figure, a and B are any two points symmetrical about the origin on the graph of function y = 1 x, AC ∥ Y axis, BC ⊥ Y axis, then the area s of △ ABC=______ .
The area of △ ABC s = 4 × 12 ×| K | = 2
If the image of quadratic function y = ax ^ 2 passes (- 1,4) and intersects with the image of linear function y = ax + 8, calculate the area of △ AOB (o is the origin of coordinates)
-1,4, take it in and get a = 4
Then y = 4x ^ 2 and y = 4x + 8 are solved to get the equation
A:x=-1,y=4;B:x=2,y=16
Then y = 4x + 8 intersects with X axis C (- 2,0)
AA ~ is perpendicular to X axis and BB ~ is perpendicular to B~
S triangle AOB = s CB ~ B-S Ca ~ A-S ob ~ b
Area of their own ~ draw a picture is obvious
Do it yourself. I'll teach you how.
By substituting the points into the matrix, the analytic expressions of the two functions are obtained. Because the two functions intersect at AB, the horizontal and vertical coordinates of the intersection are equal. Find x when y is equal. Then write a, B. Then take the X or Y axis as the dividing line to calculate the area of the two triangles. Add up. (area of 3 triangles required if necessary)
Given two points on the image of quadratic function, how to find another point and two known points to form a right triangle?
There are usually two points that match‘
Connect two known points, make a vertical line through one of them, use various relations to solve the equation of this vertical line, and combine with quadratic function to find the third point
Another point is the same
In RT triangle ABC, the angle c = 90, AC = 6, ab = 10. When r takes the following values, what is the positional relationship between a circle with C as its center and R as its radius and ab?
1.r=4 2.r=4.8 3.r=6
When r = 4, we are separated
When r = 4.8, tangent
When r = 6, intersection
Because angle c = 90, AC = 6, ab = 10
So in RT triangle BC = 8 (Pythagorean theorem)
When r = 4, the circle with C as the center and R as the radius is separated from ab
When r = 4.8, the circle with C as the center and R as the radius is tangent to ab
When r = 6, the circle with C as the center and R as the radius intersects ab
one hundred and two thousand five hundred and forty-five
The position of RT △ AOB in the plane rectangular coordinate system is shown in the figure. Point 0 is the origin, point a (0,8), point B (6.0), and point P is on the line AB, and AP = 6.1
1. Find the coordinates of point P. 2. Whether there is a point Q on the x-axis, so that the triangle with B, P, Q as the vertex is similar to △ AOB. If there is, ask for the coordinates of point Q. if not, please explain the reason
(1) Because AB: y = - 4 / 3x + 8, through P, PK is perpendicular to y axis and K, let PK be x, AK = 8 - (- 4 / 3x + 8) = 3 / 4x, according to Pythagorean theorem, x = 4.8p (4.8,3.6)
(2) Because angle ABO = angle PBQ
When bpq = 90, Pb = 4, QB = 20 / 3, q (- 2 / 3,0)
When BQP = 90, BP = 4, BQ = 2.4, q (3.6,0)
(1) According to Pythagorean theorem, ab = 10, let the coordinates of point p be (x, y),
Then there is triangle similarity, and apab = xob can be obtained. Substituting the numerical value, x = 3.6 can be obtained
The solution of ab-apab = yoa is y = 3.2
So the coordinates of point P are (3.6, 3.2)
(2) Suppose that the coordinate of Q point is (Q, 0), if BP is hypotenuse, then q = 3.6
If BQ is a hypotenuse, then bpob = bqab gives BQ = 203, because ob = 6, so q = - 23
Therefore, the coordinates of Q point are (3.6
(1) According to Pythagorean theorem, ab = 10, let the coordinates of point p be (x, y),
Then there is triangle similarity, and apab = xob can be obtained. Substituting the numerical value, x = 3.6 can be obtained
The solution of ab-apab = yoa is y = 3.2
So the coordinates of point P are (3.6, 3.2)
(2) Suppose that the coordinate of Q point is (Q, 0), if BP is hypotenuse, then q = 3.6
If BQ is a hypotenuse, then bpob = bqab gives BQ = 203, because ob = 6, so q = - 23
So the coordinates of Q point are (3.6, 0) or (- 23, 0)
(1) According to Pythagorean theorem, ab = 10, let the coordinates of point p be (x, y),
Then there is triangle similarity, and apab = xob can be obtained. Substituting the numerical value, x = 3.6 can be obtained
The solution of ab-apab = yoa is y = 3.2
So the coordinates of point P are (3.6, 3.2)
(2) Suppose that the coordinate of Q point is (Q, 0), if BP is hypotenuse, then q = 3.6
If BQ is a hypotenuse, then BP / ob = Bq / AB gives BQ = 20 / 3, because ob = 6, so q = - 2 / 3
So the expansion coordinates are q-points
(1) Let AB be the coordinate of X, y = 10,
Then there is triangle similarity, and apab = xob can be obtained. Substituting the numerical value, x = 3.6 can be obtained
The solution of ab-apab = yoa is y = 3.2
So the coordinates of point P are (3.6, 3.2)
(2) Suppose that the coordinate of Q point is (Q, 0), if BP is hypotenuse, then q = 3.6
If BQ is a hypotenuse, then BP / ob = Bq / AB gives BQ = 20 / 3, because ob = 6, so q = - 2 / 3
So the coordinates of Q point are (3.6, 0) or (- 2 / 3, 0)
The positions of line segments AB and Cd in the plane rectangular coordinate system are shown in the figure. O is the coordinate origin. If the coordinates of a point P on line segment AB are (a, b), then the coordinates of the intersection of line OP and line segment CD are (a, b)______ .
Let the intersection of line OP and line segment CD be e, ∵ ab ∥ CD, and O, B and D are on the same line, OB = BD ∥ OP = PE ∥ if the coordinates of point P are (a, b), ∥ the coordinates of point e are (2a, 2b). So the answer is (2a, 2b)
If the distance between the image vertex and the coordinate origin of the quadratic function y = x ^ 2-6x + C is 5, then C=
The answer is 13 or 5. Write down the process
Abscissa of vertex = - B / 2A = - (- 6) / 2 = 3
So ordinate = 4 or - 4
Bring in
Ordinate = (4ac-b ^ 2) / 4A = ± 4
When ordinate = 4, C = 13
When ordinate = - 4, C = 5
Let o be the coordinate origin, a (0,2), B (4,6), vector OC = λ vector OA + μ vector AB, if the vector OC ⊥ vector AB, and the area of △ ABC is 12
Finding the value of λ + μ
OC = (4 μ, 2 λ + 6 μ) vector AB = (4,4)
∴16μ+8λ+24μ=0
∴λ=-5μ
OC=(4μ,-4μ)
The angle between OC and Y axis, that is, the angle between OC and OA is 45 degrees
The distance from O to AB is the root 2
|AB|=4√2
The distance between C and ab is 12 * 2 / 4 √ 2 = 3 √ 2
So OC = 4 √ 2, and C is in the second quadrant | OC | = 4 √ 2 | μ | = 4 √ 2 μ
μ=1 λ+μ=-4
Or OC = 2 √ 2, then C is in the fourth quadrant | OC | = 4 √ 2 | μ | = - 4 √ 2 μ
μ=-1/2 λ+μ=2
Vector OC = (4 μ, 2 λ + 6 μ) vector AB = (4,4) vector OC ⊥ vector AB, 16 μ + 8 μ + 24 μ = 0. Another formula is to calculate the area according to the ABC three-point coordinates, and use the length of line AB and the distance from C to line AB to find it. It seems that there are two solutions. Let's solve it by ourselves.