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The area of a triangle can be demonstrated, for example by means of the congruence of triangles, as half of the area of a parallelogram that has the same base length and height. A graphic derivation of the formula T = h 2 b {\displaystyle T={\frac {h}{2}}b} that avoids the usual procedure of doubling the area of the triangle and then halving it.
To find the angle of a rotation, once the axis of the rotation is known, select a vector v perpendicular to the axis. Then the angle of the rotation is the angle between v and Rv. A more direct method, however, is to simply calculate the trace: the sum of the diagonal elements of the rotation matrix.
In this example, the triangle's side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well when the side lengths are real numbers. As long as they obey the strict triangle inequality, they define a triangle in the Euclidean plane whose area is a positive real number.
Just as the magnitude of a plane angle in radians at the vertex of a circular sector is the ratio of the length of its arc to its radius, the magnitude of a solid angle in steradians is the ratio of the area covered on a sphere by an object to the square of the radius of the sphere. The formula for the magnitude of the solid angle in steradians is
In two dimensions, only a single angle is needed to specify a rotation about the origin – the angle of rotation that specifies an element of the circle group (also known as U(1)). The rotation is acting to rotate an object counterclockwise through an angle θ about the origin; see below for details.
Figure 1: Euler's rotation theorem. A great circle transforms to another great circle under rotations, leaving always a diameter of the sphere in its original position. Figure 2: A rotation represented by an Euler axis and angle. In three dimensions, angular displacement is an entity with a direction and a magnitude.
If D = 1, a unique solution exists: γ = 90°, i.e., the triangle is right-angled. If D < 1 two alternatives are possible. If b ≥ c, then β ≥ γ (the larger side corresponds to a larger angle). Since no triangle can have two obtuse angles, γ is an acute angle and the solution γ = arcsin D is unique.
Since the area of a triangle cannot be negative the spherical excess is always positive. It is not necessarily small, because the sum of the angles may attain 5 π (3 π for proper angles). For example, an octant of a sphere is a spherical triangle with three right angles, so that the excess is π /2.