If the angle xoy is 60 ° and M is a point in the angle xoy, and the distance from it to ox is Ma = 2, and the distance from it to oy is MB = 11, then OM=_____ I hope it's quick and the process is detailed

If the angle xoy is 60 ° and M is a point in the angle xoy, and the distance from it to ox is Ma = 2, and the distance from it to oy is MB = 11, then OM=_____ I hope it's quick and the process is detailed


Prolonging the intersection of AM and ob at point P
∠APO=30°
MP=2MB=22
AP=22+2=24
OA = 8 change number 3
Square of OM = square of Ma + square of OA
OM=14



Given that ∠ xoy = 60 °, the distance from m to ox is me = 2, and the distance from m to oy is MF = 11


Make the vertical line of oy from point E, and the vertical point is f to get EF parallel to MF
Make the vertical line of EF from point m, and the vertical point is d to get ∠ Med = 60 ° and EM = 2, so Ed = 1
And EF = MF = 11, so Ed = 12
Therefore, OE = 24 / (root 3)
In triangle OEM, OM = 14 can be obtained by Pythagorean theorem
I can't type a lot of things when I draw the picture on the draft paper



As shown in the figure, the angle xoy = 60 degrees, P is a point within the angle xoy = 60 degrees, the distance PA from P to ox = 2, the distance Pb from P to oy = 11, and the op length is calculated


If the intersection of BP and ox is C, OCB = 30 degree, PA = 2, then PC = 4, AC = 2, root 3
BC=PB+PC=15
OC = 10 root sign 3
OA = oc-ac = 8 radical 3
OP=14



As shown in the figure, the masses of objects a and B are ma and MB respectively. They are relatively static and slide down the slope at a constant speed together. Then ()
As shown in the figure, the masses of objects a and B are ma and MB respectively. They are relatively static and slide down the slope at a constant speed together. Then ()
A. There is no static friction between a and B
B. B is affected by the sliding friction of the inclined plane
C. The inclined plane is subject to the sliding friction of B, and the direction is downward along the inclined plane
D. Dynamic friction coefficient between B and inclined plane
Question: if B has upward friction against a along the inclined plane, then B is also subject to the reaction of a friction force at the same time. I don't know how to operate this isolation method. Why can the friction force collected by a be directly used as a force of AB~


A. The acceleration is zero because of the constant speed of sliding. For a, the analysis of the force is that the gravity is vertical downward, and the supporting force of B to a is perpendicular to the direction of the inclined plane upward. These two forces are not in a straight line, so the resultant force cannot be zero, so a must be subject to the friction of B, and the friction along the direction of the inclined plane upward = = > a wrong; B, for the whole a + B, B