IPMAT Question Paper 2021 | IPM Indore Quantitative Ability

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Question 15 : If the angles A, B, C of a triangle are in arithmetic progression such that sin(2A + B) = 1/2 then sin(B + 2C) is equal to

1. \(\frac{-1}{2}\)
2. \(\frac{1}{2}\) 
3. \(\frac{-1}{\sqrt{2}}\)
4. \(\frac{3}{\sqrt{2}}\)

Video Explanation

Explanatory Answer

The angles A, B and C are in arithmetic progression and so the angles can be written as,
(B-d), B, (B+d), where d is the common difference between the angles.
The sum of the three angles of a triangle is 180 degrees and so,
B-d + B + B+d = 180^o
Or
3B = 180^o
B = 60^o

Now, we also know that sin(2A + B) = 1/2
So, 2A + B = 30^o or 150^o
But since we already know B = 60^o and the angles cannot be negative,
2A + B cannot be 30^o
2A + B = 150^o
2A + 60^o = 150^o
Or A = 45^o
=> C = 75^o

Now that we know all the angles we can find out the value of sin(B + 2C)
B + 2C = 60 + 150 = 210^o
sin(210^o) = sin(180^o + 30^o)
= sin(30^o) [sin(180^o + \(\theta\)) = - sin\(\theta\)]
= \(\frac{-1}{2}\)

The question is " If the angles A, B, C of a triangle are in arithmetic progression such that sin(2A + B) = 1/2 then sin(B + 2C) is equal to "

Hence, the answer is '\(\frac{-1}{2}\)'

Choice A is the correct answer.

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