I have not done the computations, but I assume the issue is you can take two different vectors as the normal vector for the plane, and both works. Similarly you can take a different choice for the direction vector of the line. In practice one always choose the vector pairs such that the resulting angle is less or equal to 90 degrees. In your case the actual angle should be $$ 180-102.7=77.3 $$ and as a result the angle between the plane and the line should be $$ 90-77.3=12.7 $$ instead.
1. Find radians reflection angle?
The relevant quantity here is the velocity vector of the ball. When you hit a wall, you want to flip the component of that vector perpendicular to the wall and keep the other component equal.So, if you fix a velocity vector by giving an initial speed $v_0$ and a direction given by an angle $alpha$ with respect to the horizontal direction and assuming there are only horizontal end vertical walls, this is what will happen:.
2. If one of the angle of a triangle is 110 degree, then what is the angle between the bisector of the other two angles?
remaining angles sum to 180-110=70 degrees. The bisectors of the other two angles will capture between those two internal angles for the triangle so formed, 70/2 = 35 degrees. This will leave 180-35=145 degrees for the angle formed between these two bisectors. If one of the angle of a triangle is 110 degree, then what is the angle between the bisector of the other two angles?
3. Is kurt angle really that up himself?
Why can not kurt stop being a little pussy, if he does not want other people using the same moves he does, why does not he just go back to WWE, cause TNA sucks anyway
4. angle between two vector ?
km do not "convert" to degrees. suppose you had an equilateral triangle with sides =34 km each. the angles would be 60 degrees, right? as long as the three equal side measurements are in the same units, you know the angle. Likewise with your vectors: if all components are measured in the same units, you can ignore the units when computing the angle. So.... if u = 6i 4j v= 2i 7j and C is the angle between them, then we know u dot v = |u| |v| cos C u dot v = 12 28 = 40 |u| = sqrt( 36 16) = sqrt(52) = 2 sqrt(13) |v| = sqrt( 4 49) = sqrt( 53) so cos C = 40/(2sqrt(13) * sqrt(53)) = 20/(sqrt(13)*sqrt(53)) C is the inverse cosine of 20/(sqrt(13)*sqrt(53)) --------------------
5. Calculating View Angle?
For accurate calculations, convert (lat, lon, elevation) directly to earth-centered (x,y,z). (If you do not do this, you need to retain additional information about the local normal ["up"] directions in order to compute angles accurately at nonzero elevations.)Given two points (x,y,z) and (x',y',z') in an earth-centered coordinate system, the vector from the first to the second is (dx,dy,dz) = (x'-x, y'-y, z'-z), whence the cosine of the angle made to the normal at (x,y,z) is the inner product of the unit length versions of those vectors:Obtain its principal inverse cosine. Subtract this from 90 degrees if you want the angle of view relative to a nominal horizon. This is the "elevation."A similar calculation obtains the local direction of view ("azimuth"). We need a level vector (u,v,w) that points due north. One such vector at the location (x,y,z) is (-zx, -zy, x^2y^2). (The inner product of these two vectors is zero, proving it is level. Its projection onto the Equatorial plane is proportional to (-x,-y) which points directly inward, making it the projection of a north-pointing vector. These two calculations confirm that this is indeed the desired vector). ThereforeWe also need the sine of the azimuth, which is similarly obtained once we know a vector pointing due East (locally). Such a vector is (-y, x, 0), because it clearly is perpendicular to (x,y,z) (the up direction) and the northern direction. ThereforeThese values enable us to recover the azimuth as the inverse tangent of the cosine and sine.A pilot in an airplane flying west at 4000 meters, located at (lat, lon) = (39, -75), sees a jet far ahead located at (39, -76) and flying at 12000 meters. What is are the angles of view (relative to the level direction at the pilot's location)?The XYZ coordinates of the airplanes are (x,y,z) = (1285410, -4797210, 3994830) and (x',y',z') = (1202990, -4824940, 3999870), respectively (in the ITRF00 datum, which uses the GRS80 ellipsoid). The pilot's view vector therefore is (dx,dy,dz) = (-82404.5, -27735.3, 5034.56). Applying the formula gives the cosine of the view angle as 0.0850706. Its inverse cosine is 85.1199 degrees, whence the elevation is 4.88009 degrees: the pilot is looking up by that much. A north-pointing level vector is (-5. 13499, 19. 1641, 24. 6655) (times 10^12) and an east-pointing level vector is (4.79721, 1.28541, 0) (times 10^6). Therefore, applying the last two formulas, the cosine of the azimuth equals 0.00575112 and its sine equals -0.996358. The ArcTangent function tells us the angle for the direction (0. 00575112, -0. 996358) is 270. 331 degrees: almost due west. (It's not exactly west because the two planes lie on the same circle of latitude, which is curving toward the North pole: see Why is the 'straight line' path across continent so curved? for an extended explanation.)By the way, this example confirms we got the orientation correct for the azimuth calculation: although it was clear the east-pointing vector was orthogonal to the other two vectors, until now it was not plain that it truly points east and not west