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Thales's theorem can also be used to find the centre of a circle using an object with a right angle, such as a set square or rectangular sheet of paper larger than the circle. [7] The angle is placed anywhere on its circumference (figure 1). The intersections of the two sides with the circumference define a diameter (figure 2).
The measure of ∠AOB, where O is the center of the circle, is 2α. The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that intercepts the same arc on the circle. Therefore, the angle does not change as its vertex is moved to different positions on the circle.
Trigonometric ratios can also be represented using the unit circle, which is the circle of radius 1 centered at the origin in the plane. [37] In this setting, the terminal side of an angle A placed in standard position will intersect the unit circle in a point (x,y), where = and = . [37]
The Maharashtra State Board of Secondary and Higher Secondary Education (Abbreviation: MSBSHSE) is a statutory and autonomous body established under the "Maharashtra Secondary Boards Act" 1965 (amended in 1977). [1] Most important task of the board, among few others, is to conduct the SSC for 10th class and HSC for 12th class examinations. [2]
The nine-point circle is tangent to the incircle and excircles. In geometry, the nine-point circle is a circle that can be constructed for any given triangle. It is so named because it passes through nine significant concyclic points defined from the triangle. These nine points are: [28] [29] The midpoint of each side of the triangle; The foot ...
However, if we only consider triangles whose sides are minor arcs of great circles, we have the following properties: The angle sum of a triangle is greater than 180° and less than 540°. The area of a triangle is proportional to the excess of its angle sum over 180°. Two triangles with the same angle sum are equal in area.
Determination of a circle, that intersects four circles by the same angle. [2] Solving the Problem of Apollonius; Construction of the Malfatti circles: [3] For a given triangle determine three circles, which touch each other and two sides of the triangle each. Spherical version of Malfatti's problem: [4] The triangle is a spherical one.
Symmetrically, form a circle through points B and C, tangent to edge AC, and a circle through points A and C, tangent to edge AB. These three circles have a common point, the first Brocard point of ABC. See also Tangent lines to circles. The three circles just constructed are also designated as epicycles of ABC. The second Brocard point is ...