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Using the usual notations for a triangle (see the figure at the upper right), where a, b, c are the lengths of the three sides, A, B, C are the vertices opposite those three respective sides, α, β, γ are the corresponding angles at those vertices, s is the semiperimeter, that is, s = a + b + c / 2 , and r is the radius of the inscribed circle, the law of cotangents states that
A triangle with sides a, b, and c. In geometry, Heron's formula (or Hero's formula) gives the area of a triangle in terms of the three side lengths , , . Letting be the semiperimeter of the triangle, = (+ +), the area is [1]
The area of a triangle can be demonstrated, for example by means of the congruence of triangles, as half of the area of a parallelogram that has the same base length and height. A graphic derivation of the formula T = h 2 b {\displaystyle T={\frac {h}{2}}b} that avoids the usual procedure of doubling the area of the triangle and then halving it.
Use Napier's rules to solve the triangle ABD: use c and B to find the sides AD and BD and the angle ∠BAD. Then use Napier's rules to solve the triangle ACD: that is use AD and b to find the side DC and the angles C and ∠DAC. The angle A and side a follow by addition.
The shaded blue and green triangles, and the red-outlined triangle are all right-angled and similar, and all contain the angle . The hypotenuse B D ¯ {\displaystyle {\overline {BD}}} of the red-outlined triangle has length 2 sin θ {\displaystyle 2\sin \theta } , so its side D E ¯ {\displaystyle {\overline {DE}}} has length 2 sin 2 θ ...
The law of sines (also known as the "sine rule") for an arbitrary triangle states: [85] = = = =, where is the area of the triangle and R is the radius of the circumscribed circle of the triangle:
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Similar right triangles illustrating the tangent and secant trigonometric functions Trigonometric functions and their reciprocals on the unit circle. The Pythagorean theorem applied to the blue triangle shows the identity 1 + cot 2 θ = csc 2 θ, and applied to the red triangle shows that 1 + tan 2 θ = sec 2 θ.