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If A,B,C,D are thesmallest positive angles in ascending order of mag. which have their sines equal to the positive quantities K the values of 4sinA/2 +3sinB/2 +2sinC/2 +sin D/2?

Abhinav Acharya , 10 Years ago
Grade 11
anser 2 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans: 5K
Sol:
A, B, C & D are vey small angles which means they all are very close to zero.
sinA = sinB = sinC = sinD = K
sinA = 2.sin\frac{A}{2}.cos\frac{A}{2}
sin\frac{A}{2}=\frac{K}{2.cos\frac{A}{2}}
Since A is very very close to zero, then
cos\frac{A}{2} = 1
sin\frac{A}{2} = \frac{K}{2}
Similarly for the other angles,
sin\frac{A}{2}=sin\frac{B}{2}=sin\frac{C}{2}=sin\frac{D}{2} = \frac{K}{2}
4sin\frac{A}{2}+3sin\frac{B}{2}+2sin\frac{C}{2}+sin\frac{D}{2} = 4.\frac{K}{2}+3.\frac{K}{2}+2.\frac{K}{2}+\frac{K}{2}
=5K
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

Abhinav Acharya

Last Activity: 10 Years ago

But the answer given in the book is 2 under root 1 + k

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