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
To determine the type of triangle ABC based on the given equation c(a+b) Cos(B/2) = b(c+a) Cos(C/2), we need to understand the properties of triangles and how to use trigonometric identities.
Here's how we can approach the problem step by step:
Step 1: Understand the equation
The given equation involves the lengths of the sides of the triangle (a, b, and c) and the cosine of half of each angle (B/2 and C/2). It equates the left side of the equation to the right side.
Step 2: Use the Law of Cosines
Since the equation involves the cosine of half angles, let's rewrite it using the Law of Cosines for triangles.
The Law of Cosines states that: c^2 = a^2 + b^2 - 2ab * cos(C)
Let's substitute the values from the given equation into the Law of Cosines:
c(a+b)Cos(B/2) = b(c+a)Cos(C/2)
=> c(a+b)Cos(B/2) = b(a+c)Cos(B/2)
=> c(a+b) = b(a+c)
Step 3: Simplify the equation
Expand both sides of the equation:
ac + bc + ab + cb = ab + bc + ac
Cancel out similar terms:
ac + bc = ac + bc
Step 4: Analyze the simplified equation
The simplified equation implies that the terms on both sides of the equation are equal. This means that the sides are proportional, which can happen in two cases:
Case 1: When a = c (isosceles triangle)
In an isosceles triangle, two sides are congruent. So, if a = c, the equation will be true.
Case 2: When a + c = 0 (right-angled triangle)
In a right-angled triangle, one angle is equal to 90 degrees. This implies that either cos(B/2) or cos(C/2) is equal to zero. Therefore, if a + c = 0, the equation will be true.
Step 5: Determine the type of triangle
Based on the analysis, the equation c(a+b)Cos(B/2) = b(a+c)Cos(C/2) will be true when the triangle is either isosceles or right-angled. Therefore, the correct answer is option 3: isosceles or right-angled.