If in a triangle ABC,
c(a+b) Cos B/2 =b(c+a) Cos C/2
then the triangle is :-
1. isosceles
2. Right angled
3. isosceles or right angled
4. none of these
I want the explanation
Isosceles
To determine the nature of triangle ABC based on the given equation, let's go step by step:
First, let's simplify the given equation using the trigonometric identity:
cos(B/2) = √[(s - a)(s - b)] / √(ab)
cos(C/2) = √[(s - c)(s - a)] / √(ac)
where s = (a + b + c)/2 is the semi-perimeter of triangle ABC.
Substituting the above values in the given equation:
c(a + b) √[(s - a)(s - b)] / √(ab) = b(c + a) √[(s - c)(s - a)] / √(ac)
To simplify this equation further, let's remove the common terms:
(a + b) √[(s - a)(s - b)] = (c + a) √[(s - c)(s - a)]
Squaring both sides, we get:
(a + b)²(s - a)(s - b) = (c + a)²(s - c)(s - a)
Expanding both sides, we get:
(a² + 2ab + b²)(s² - (a + b)s + ab) = (c² + 2ac + a²)(s² - (c + a)s + ac)
Simplifying further:
a²s² - a²s(a + b) + a³b + 2ab²s - 2ab² + b²s² - b²s(a + b) + ab³ = c²s² - c²s(a + b) + a³c + 2ac²s - 2ac² + a²c²
Combining like terms:
(a² - b²)s² + (2ab - 2ab)s + (a³b - ab³) + (a²c - a²c) = (c² - a²)s² + (2ac - 2ac)s + (a³c - c³a)
Simplifying and rearranging terms:
(a + b)s² = (c - a)(c + a)s
Dividing by (c - a) on both sides, we get:
(a + b)s = (c + a)s
Canceling out s on both sides, we get:
a + b = c + a
Simplifying further:
b = c
Therefore, from the given equation, we can conclude that the triangle ABC is isosceles (option 1).
Hence, the correct answer is 1. isosceles.