What is the maximum
possible value of 21Sinx+72Cosx?

(1)21 (2) 57
(3) 63 (4) 75 (5) None of these

Solution follows here:

__Solution:__
It is a direct
trigonometric formula based one.

Maximum value of
aSinx+bCosx = √(a

^{2}+b^{2})
√(a

^{2}+b^{2}) = √(21^{2}+72^{2}) = √(3^{2}7^{2}+3^{2 }24^{2}) = 3 √(7^{2}+24^{2})
= 3 √(49+576) = 3 √625 =
3*25 = 75

**Answer (4)**
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