Ads
related to: right triangle circumscribing a circle equation solver answer
Search results
Results From The WOW.Com Content Network
Hence, given the radius, r, center, P c, a point on the circle, P 0 and a unit normal of the plane containing the circle, ^, one parametric equation of the circle starting from the point P 0 and proceeding in a positively oriented (i.e., right-handed) sense about ^ is the following:
Such a circle is said to circumscribe the points or a polygon formed from them; such a polygon is said to be inscribed in the circle. Circumcircle, the circumscribed circle of a triangle, which always exists for a given triangle. Cyclic polygon, a general polygon that can be circumscribed by a circle. The vertices of this polygon are concyclic ...
Microsoft Math contains features that are designed to assist in solving mathematics, science, and tech-related problems, as well as to educate the user. The application features such tools as a graphing calculator and a unit converter. It also includes a triangle solver and an equation solver that provides step-by-step solutions to each problem.
A polygon whose vertices are concyclic is called a cyclic polygon, and the circle is called its circumscribing circle or circumcircle. All concyclic points are equidistant from the center of the circle. Three points in the plane that do not all fall on a straight line are concyclic, so every triangle is a cyclic polygon, with a well-defined ...
In trigonometry, the law of sines, sine law, sine formula, or sine rule is an equation relating the lengths of the sides of any triangle to the sines of its angles. According to the law, = = =, where a, b, and c are the lengths of the sides of a triangle, and α, β, and γ are the opposite angles (see figure 2), while R is the radius of the triangle's circumcircle.
The recursion terminates when P is empty, and a solution can be found from the points in R: for 0 or 1 points the solution is trivial, for 2 points the minimal circle has its center at the midpoint between the two points, and for 3 points the circle is the circumcircle of the triangle described by the points.
We denote further D = c / b sin β (the equation's right side). There are four possible cases: If D > 1, no such triangle exists because the side b does not reach line BC. For the same reason a solution does not exist if the angle β ≥ 90° and b ≤ c. If D = 1, a unique solution exists: γ = 90°, i.e., the triangle is right-angled.
Proposition one states: The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference of the circle. Any circle with a circumference c and a radius r is equal in area with a right triangle with the two legs being c and r.