# Prove that area of a triangle can be found by taking 1/2 of the product of the lengths of 2 sides x sine of the included angle using the figure...

Prove that area of a triangle can be found by taking 1/2 of the product of the lengths of 2 sides x sine of the included angle using the figure given

http://www.mathalino.com/sites/default/files/images/derivation-radius-circumcircle.jpg

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The premise of the area (A) of a triangle is: 1/2 x base x height. The reason for the 1/2 is that the triangle essentially represents 1/2 of a rectangle. It is imperative to represent the height of the triangle as being from it's base to the apex of the highest point on the triangle (height is not measured as a length of one of the sides, unless it is right triangle).

For simplicity, let's assume an equilateral triangle with three included angles of 60 degrees and three equal length sides of "a".

The height of the triangle, from base to heights point, is based on extending a line from the mid-point of the base to the angle at the vertex of the triangle. For example:

sin 60 = opposite / hypotonuse. In this context, opposite is the height of the triangle and the hypotonuse is "a" the leg of the triangle.

Therefore, height of the triangle is equal to: sin 60 * a. Now, referring back to the area of a triangle, with the newly acquired height we have,

A = 1/2 * base * height = 1/2 * a * (sin 60 * a)

From this you can easily deduce that the area of a triangle is equal to:

(a*a)/2 * sin 60 or to generalize: (a * a)/2 * sin`Theta`

`` This provides the proof.