Search results
Results from the WOW.Com Content Network
Classically the defect arises in two contexts: in the Euclidean plane, angles about a point add up to 360°, while interior angles in a triangle add up to 180°. However, on a convex polyhedron , the angles of the faces meeting at a vertex add up to less than 360° (a defect), while the angles at some vertices of a nonconvex polyhedron may add ...
A method to solve such problems is to consider the rate of change of the angle in degrees per minute. The hour hand of a normal 12-hour analogue clock turns 360° in 12 hours (720 minutes) or 0.5° per minute. The minute hand rotates through 360° in 60 minutes or 6° per minute. [1]
Solution of triangles (Latin: solutio triangulorum) is the main trigonometric problem of finding the characteristics of a triangle (angles and lengths of sides), when some of these are known. The triangle can be located on a plane or on a sphere. Applications requiring triangle solutions include geodesy, astronomy, construction, and navigation.
An easy formula for these properties is that in any three points in any shape, there is a triangle formed. Triangle ABC (example) has 3 points, and therefore, three angles; angle A, angle B, and angle C. Angle A, B, and C will always, when put together, will form 360 degrees. So, ∠A + ∠B + ∠C = 360°
Sometimes it is desirable to have a triangulation with special properties, e.g., in which all triangles have large angles (long and narrow ("splinter") triangles are avoided). [3] Given a set of edges that connect points of the plane, the problem to determine whether they contain a triangulation is NP-complete. [4]
As can be seen from Fig. 1, these problems involve solving the triangle NAB given one angle, α 1 for the direct problem and λ 12 = λ 2 − λ 1 for the inverse problem, and its two adjacent sides. For a sphere the solutions to these problems are simple exercises in spherical trigonometry , whose solution is given by formulas for solving a ...
The time required for a vehicle to enter a controlled intersection from a stop is the sum of the perception time (t p), the time required to actuate an automatic transmission or shift to first gear (t c), and the time to accelerate and enter or traverse the road (t a). The sum of the first two quantities is t pc.
The diagrams we use show this construction for an acute angle, but it indeed works for any angle up to 180 degrees. This requires three facts from geometry (at right): Any full set of angles on a straight line add to 180°, The sum of angles of any triangle is 180°, and,