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
1. Isosceles
Put B=C
AFTER PUTTING AND SIMPLIFYING WE ARE GETTING AB=AC THEN TWO SIDES ARE EQUAL THEN IT IS A ISOSCELES TRIANGLE
To determine the type of triangle, we need to understand the properties given in the equation:
c(a+b) * cos(B/2) = b(c+a) * cos(C/2)
To analyze this equation, let's break it down step by step and see if we can determine any properties of the triangle based on it:
1. Expand both sides of the equation:
ca * cos(B/2) + cb * cos(B/2) = bc * cos(C/2) + ba * cos(C/2)
2. Rearrange the terms:
ca * cos(B/2) - bc * cos(C/2) = ba * cos(C/2) - cb * cos(B/2)
3. Notice that the left side of the equation can be rewritten using the cosine difference formula:
2 * (ca * sin((B + C)/2) * sin((B - C)/2)) = 2 * (ba * sin((B + C)/2) * sin((B - C)/2))
Canceling out the common terms, we get:
ca * sin((B - C)/2) = ba * sin((B - C)/2)
4. Divide both sides of the equation by sin((B - C)/2) (assuming sin((B - C)/2) is not zero):
ca = ba
From step 4, we can conclude that side AC is equal to side AB (ca = ba). Hence, the triangle ABC is an isosceles triangle.
Therefore, the correct answer is 1. isosceles.