# Trigonometry of Triangles

## I will discuss in this articel the trigonometric properties of triangles.

Triangles have unque trigonometric properties just like other geometrical figures. The inside circle of a triangle is constructed by bisecting the triangle angles and then taking the perpendicular distance from the three bisectors and drawing the inside circle. The Circumscribed circle is constructed by bisecting the sides of the triangle and taking the distance of the three side bisectors and the corners of the triangle as radius. The radius of the inside and outside radius is related mathematically to the dimensions of any triangle.

In any triangle, the tangent of the inside circle is perpendicular to the radius. In addition, the angle bisectors meet at the centre of the inside circle. That is, tan (A/2) = r/ AX, where Ox is the distance from the triangle angle (A) to the radius meeting the side AB or the side against angle C or it can be labeled (c). In the same way Tan (B/2) = r/Bx. That is, AX = r/Tan(A/2) and BX = r/Tan B/2. However AB = AX + BX. That is, (c) = r * [ 1/Tan (A/2) +/ Tan B/2]. In other words r = c* ( Tan A/2 + Tan B/2). In the same manner, r = b* ( Tan A/2 + Tan C/2) or r = a* ( Tan B/2 + Tan C/2). That for any triangle (tan A/2 + Tan B/2) / ( Tan A/2 + Tan C/2) = b/c and (tan A/2 + Tan C/2)/ Tan B/2 + Tan C/2) = a/b. Applying the sin theorem a/ Sin A = b/ Sin B. That is, b Sin A = a Sin B, there fore a/b = Sin A/ Sin B in the same manner, b/c = Sin B/ Sin C . That is, the inside circle radius can be expressed in terms of a side length and Sine ratio of two angles and tan of half of the two angles as follows:

r = c* b/c * ( Tan A/2 + Tan C/2) = c* Sin B/Sin C*( Tan A/2 + Tan C/2)

r = b*c/b* ( Tan A/2 + Tan B/2) = b* Sin C/ Sin B* ( Tan A/2 + Tan B/2)

r = a* b/a * ( Tan A/2 + Tan C/2) = a* Sin B/Sin A* ( Tan C/2 + Tan A/2)

Circumscribed radius of circle of any triangle

The angle two radius make at the circumference of a circle s half of the anle they make at the center of a circle. In this respect, if one bisects the sides of a circle to draw a circumscribed circle of any triangle, then the angle it makes at the center is twice the angle of the one of the triangle. In addition the perpendicular from the center of a circle bisects the cord or one side of any triangle. That is, The angle between the radius of the circumscribed  circle and the perpendicular is equal to the opposite angle of the side. Say the angle is (A) and the opposite side is (a) then Sin A = a/2/ R where R is the radius of the circumscribed circle of a triangle. In other words, R Sin A = a/2. That is R = a/2/ Sin A = a/s*co sec A or R = a/2/ 2 Sin A/2* cos A/2 = a/4 sin A/2* Cos A/2.

R = a/ 4 Sin A/2* Cos A/2

R = b/ 4 Sin B/2 8 Cos A/2

R = c/ 4 Sin C/2* Cos C/2

That is if one knows the length of any side of any triangle and knows one angle and its trigonometric identity , that is sin of any angle then one can calculate the length of the radius of the circumscribed circle of a triangle.For example the radius of the circumscribed circle of an equilateral triangle of length of 4 cm is equal to 4/  4Sin30*cos30. Sin30 = 0.5 and cos 30 = square toot 3/2. There fore R = 4/ 4*0.5*square rootof 3/2 = 2 /0.5*square root of 3 = 4/ square root of 3.

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