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The sides of the triangle emanating from the North Pole (great circles of the sphere) both meet the equator at right angles, so this triangle has an exterior angle that is equal to a remote interior angle. The other interior angle (at the North Pole) can be made larger than 90°, further emphasizing the failure of this statement.
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 ...
Exterior angles can be also defined, and the Euclidean triangle postulate can be formulated as the exterior angle theorem. One can also consider the sum of all three exterior angles, that equals to 360° [9] in the Euclidean case (as for any convex polygon), is less than 360° in the spherical case, and is greater than 360° in the hyperbolic case.
An exterior angle of a triangle is an angle that is a linear pair (and hence supplementary) to an interior angle. The measure of an exterior angle of a triangle is equal to the sum of the measures of the two interior angles that are not adjacent to it; this is the exterior angle theorem. [34]
English: This file illustrates the special right triangle of angles 45°, 45° and 90°. Inside, the triangle is depicted with all of its angles and side lengths. The right angle is marked as a small square. To the right of the 90° angle is another angle represented by an arc, and its measure, 45°, is written to the left of it. Atop the right ...
The values of the trigonometric functions can be evaluated exactly for certain angles using right triangles with special angles. These include the 30-60-90 triangle which can be used to evaluate the trigonometric functions for any multiple of , and the isosceles right triangle or 45-45-90 triangle which can be used to evaluate the trigonometric ...
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.
The formula can be proved by using mathematical induction: starting with a triangle, for which the angle sum is 180°, then replacing one side with two sides connected at another vertex, and so on. The sum of the external angles of any simple polygon, if only one of the two external angles is assumed at each vertex, is 2π radians (360°).