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An equilateral triangle is a triangle in which all three sides have the same length, and all three angles are equal. Because of these properties, the equilateral triangle is a regular polygon, occasionally known as the regular triangle. It is the special case of an isosceles triangle by modern definition, creating more special properties.
Triangles have many types based on the length of the sides and the angles. A triangle whose sides are all the same length is an equilateral triangle, [3] a triangle with two sides having the same length is an isosceles triangle, [4] [a] and a triangle with three different-length sides is a scalene triangle. [7]
In an equilateral triangle, the 3 angles are equal and sum to 180°, therefore each corner angle is 60°. Bisecting one corner, the special right triangle with angles 30-60-90 is obtained. By symmetry, the bisected side is half of the side of the equilateral triangle, so one concludes sin ( 30 ∘ ) = 1 / 2 {\displaystyle \sin(30^{\circ ...
Set square shaped as 45° - 45° - 90° triangle The side lengths of a 45° - 45° - 90° triangle 45° - 45° - 90° right triangle of hypotenuse length 1.. In plane geometry, dividing a square along its diagonal results in two isosceles right triangles, each with one right angle (90°, π / 2 radians) and two other congruent angles each measuring half of a right angle (45°, or ...
A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation 4x 3 − 3x + d = 0, where is the value of the cosine function at the one-third angle and d is the known value of the cosine function at the full angle.
Langley's Adventitious Angles Solution to Langley's 80-80-20 triangle problem. Langley's Adventitious Angles is a puzzle in which one must infer an angle in a geometric diagram from other given angles. It was posed by Edward Mann Langley in The Mathematical Gazette in 1922. [1] [2]
Given a square, construct equilateral triangles on two adjacent edges, either both inside or both outside the square. Then the triangle formed by joining the vertex of the square distant from both triangles and the vertices of the triangles distant from the square is equilateral. [2]
Let A'BC be the equilateral triangle having base BC and vertex A' on the negative side of BC and let AB'C and ABC' be similarly constructed equilateral triangles based on the other two sides of triangle ABC. Then the lines AA', BB', CC' are concurrent and the point of concurrence is the 1st isogonal center. Its trilinear coordinates are