As shown in the figure, if ∠ ABC is known, draw the figure after ∠ ABC rotates 50 ° clockwise around point o

As shown in the figure, if ∠ ABC is known, draw the figure after ∠ ABC rotates 50 ° clockwise around point o

The graph is as follows

I'm not very good at it. I'm going to draw a picture of the triangle ABC after it rotates 120 degrees clockwise around point O. I need to draw a trace

First connect Ao, then take ao as radius, O as center, draw circle clockwise, measure 120 ° angle from Ao clockwise, make straight line OA ', the intersection of OA' and arc is the point a rotates 120 ° clockwise around o point, draw B ', C' of B and C rotate 120 ° around o point in turn, connect a ', B', C ', get △ a'b'c' is the figure that △ ABC rotates 120 ° clockwise around o point

In triangle ABC, if a satisfies √ 3sina + cosa = 1, ab = 2, BC = 2 √ 3, then the area of triangle ABC is

In triangle ABC, if a satisfies √ 3sina + cosa = 1, ab = 2, BC = 2 √ 3, then the area of triangle ABC is
√3sinA+cosA=1
Sin (a + 30 degrees) = 1 / 2
A = 120 degrees
sinC=sinA*AB/BC=1/2
C = 30 degrees
AC=AB=2
Area of triangle ABC = √ 3

In △ ABC, given the edge C = 10 and cosacosb = Ba = 43, find the radius of inscribed circle of a, B and △ ABC

It is known that the length of one side of the isosceles triangle ABC is 4, and the other two sides are the two real roots of the equation x ^ 2-12x + M = O about X,
Finding the value of M and the circumference of isosceles triangle ABC

If the other two sides are waist, then the two roots of the equation are equal,
So, 12 * 12-4 * m = 0
So, M = 36,
The solution of the equation is X1 = x2 = 6
The circumference of an isosceles triangle is 16
If one of the other two sides is not waist, then the length of one side is 4, then 4 is a root of the equation,
So m = 32
The other root of the equation is 8. In this case, 4,4,8 cannot form a triangle,
So m = 32 does not meet the requirements

In △ ABC, cosa = 3 / 5, SINB = 5 / 13, find COSC=

cosA=3/5
sinA=4/5
sinB=5/13
0

In △ ABC, cosa = 5 / 13, SINB = 3 / 5, then COSC =?
I figure out that the answer is 16 / 65 or 56 / 65. Why can I give up 56 / 65?

Because it's a triangle, the angle corresponding to the number 180 is too big, so it's abandoned

The focal length of the ellipse is 6 and passes through point P (4125)

Let the equation of ellipse be x2a2 + y2b2 = 1 (a ＞ B ＞ 0), the focal length of ellipse be 6, and pass through the point P (4125), ｜ 2A2 − B2 = 642a2 + (125) 2B2 = 1, the solution is a & nbsp; = 5B = 4 (negative rounding). Therefore, the equation of ellipse is X225 + y216 = 1

It is known that the focus of the ellipse is on the x-axis, the length of the long axis is 12, and the focal length is 8

According to the meaning of the title,
2a=12,2c=8
A = 6, C = 4
b^2=a^2 --c^2=20
So the elliptic standard equation is x ^ 2 / 36 + y ^ 2 / 20 = 1

If the sum of major axis and minor axis of the ellipse with focus on Y axis is 18, and the focal length is 6, then the standard equation of the ellipse is (how to calculate?)

Since the focus is on the Y axis, let the equation of the ellipse be y ^ 2 / A ^ 2 + x ^ 2 / b ^ 2 = 1 (a > b > 0)
The focal length is 6, which means 2C = 6 and C = 3
Then think of the relationship between the major axis and minor axis of the ellipse and the focal length, a ^ 2-B ^ 2 = C ^ 2 = (3) ^ 2 = 9
The sum of major axis and minor axis is 18, which means 2A + 2B = 18
At the same time, a = 5 and B = 4 are obtained
Then the equation of ellipse is y ^ 2 / 25 + x ^ 2 / 16 = 1