In any angle ABC prove that the perimeter = a/Sin A(Sin A + Sin B + Sin C)
Start with perimeter = a+b+c
but b= aSinB/sinA and
c= aSinC/SinA
and if you don't remember, a=a=a*SinA/
SinA
To prove that the perimeter of any triangle ABC is equal to a/Sin A (Sin A + Sin B + Sin C), we will start with the formula for the perimeter of a triangle.
The perimeter of a triangle ABC is defined as the sum of the lengths of its three sides, denoted as a, b, and c.
So initially, we have perimeter = a + b + c.
Now, we will substitute the values of b and c using trigonometric ratios.
In a triangle, each angle can be related to its opposite side using the sine function. This relationship is given by the sine rule or law of sines.
According to the law of sines, for any angle A in triangle ABC,
a/Sin A = b/Sin B = c/Sin C
Using this relation, we can rewrite b and c in terms of a and the corresponding angles:
b = a * Sin B / Sin A
c = a * Sin C / Sin A
Substituting these values back into the perimeter formula, we get:
perimeter = a + (a * Sin B / Sin A) + (a * Sin C / Sin A)
Now we can simplify the expression:
perimeter = a * [1 + (Sin B / Sin A) + (Sin C / Sin A)]
Since Sin B / Sin A and Sin C / Sin A are ratios of sine functions, we can rewrite them using the identities:
Sin B / Sin A = Sin A / Sin A * Sin B / Sin A = Sin A * Cot A * Sin B * Cot A = (Sin A * Sin B) / (Sin A * Sin A) = Sin B / Sin A
Similarly, Sin C / Sin A = Sin C / Sin A
Now substituting back into the expression:
perimeter = a * [1 + (Sin B / Sin A) + (Sin C / Sin A)]
= a * [1 + Sin B / Sin A + Sin C / Sin A]
= a * [Sin A / Sin A + Sin B / Sin A + Sin C / Sin A]
= a * [Sin A (1 / Sin A + Sin B / Sin A + Sin C / Sin A)]
= a * [Sin A (Sin A + Sin B + Sin C) / Sin A]
= a * (Sin A + Sin B + Sin C)
Therefore, we have proved that the perimeter of any triangle ABC is equal to a/Sin A (Sin A + Sin B + Sin C).