The sum of the angles of a triangle is 180.
Hence we can say that:
a+b+c=180

Now we need to find all possible cases when:
a-b+c<0 or

a+c<b

Thing to note: A and C has to be at least 1.
Hence,
A+C=180-b

Hence:
B>90

But B cannot be 180 or 179 as A and c have to be atleast 1.
Hence the maximum value of B is 178

Total possible values:
178-90=88

Hence answer should be A.

For a triangle the sum of three sides a+b+c=180

since here in we need to find "the number of possible values of a – b + c, which are less than zero"

or say a+c<b and (a+c)-b <0 ;
since c=180-a-b

a – b + c, substitute value of c that a-b+180-a-b
=> -2b=-180
=> b=90

so if b=90 , then a+c has to be 90 , and value of a,c can be any thing (89,1), (81,9) ... but since we need to find the number of possible values of a – b + c, which are less than zero ,value of b has to be 91 and the values of a & c can be anything from (88,1) , (1,88) total values such values which we would fetch us sum < 0 would be IMO 88 ; so IMO A is correct..

Solution

Given:

• We are given that a, b and c are the interior angles of a triangle ABC
• a, b and c are integers

To find:

• The number of possible values of a – b + c, which are < 0

Approach and Working:
Since, a, b and c, are interior angles of a triangle, we can say that,

• a + b + c = \(180^o\)
o Implies, a + c = 180 – b
• Substituting this in the expression, a – b + c, we get,
o a – b + c = (a + c) – b = (180 – b ) – b = 180 – 2b

For the value of 180 – 2b to be < 0

• 2b > 180
• b > 90

And, the maximum value that ‘b’ can take is \(178^o\), since, the minimum value of a and c are \(1^o\) each.

• Thus, 91 ≤ b ≤ 178

Therefore, the total number of values that b can take is 88

Hence the correct answer is Option A.

Answer: A


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