A simple problem for a simple day
Angles and belong to a triangle and satisfy the following:
Sumit your solution in form: .
Function has a value of remainder when is divided by .
e. g. ,
Inspiration: Solution section.
The answer is 146.
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Observatory Solution 8 3 − 3 = ( 2 1 ) ( 2 2 3 − 1 ) ( 2 3 ) = ( sin 4 5 ) ( sin 1 5 ) ( s i n 1 2 0 ) Using the second equation we can easily verify α = 4 5 , β = 1 5 and hence γ = 1 2 0 Formal Solution Using sin2A+sin2B+sin2C=4sinA sinB sinC given A+B+C= π sin 2 α + sin 2 β + sin 2 γ = 4 sin α × sin β × sin γ = 4 8 3 − 3 ⇒ 3 2 + sin γ = 2 3 − 3 ⇒ γ = 1 2 0 and solving the equations we get the previously obtained values i.e a=45, b=15 and c=120. R(a,17)=11,R(b,17)=15. Hence, 1 2 0 + 1 1 + 1 5 = 1 4 6