Solve this: In a triangle ABC (a) sinA sinB sinC ≤ 338 - Maths - Straight Lines

Question

Solve this:
In a triangle ABC
( a )   sin A   sin B   sin C   ≤  
3 3 8                
                 
    ( b )   sin 2 A + sin 2 B + sin 2 C ≤ 9 4
( c )   sin A   sin B   sin C   i s   a l
w a y s   p o s i t i v e         ( d )
  sin 2 A + sin 2 B ≤ 1 + cos   C

Dipak Yugnirmal · Accepted Answer

Dear Student

Let Z = sinA sinB sinC

= sinA sinB sin[π - (A+B)] [ In any triangle , A+B +C = π
]

= SinA sinB sin(A+B)

= sinA sinB (sinA cosB + cosA sinB)

= sin^2 A sinB cosB + sin^2 B sinA cosA

To find the maximum value,

δZ/δA = 0

= 2sinA cosA sinB cosB + sin^2 B * cos2A = 0

= sin2A cosB + cos2A sinB = 0

= sin(2A + B) = 0

= 2A + B = π ( i )

Similarly, 
δZ/δB = 0

=> 2B + A = π ( ii )

Solving ( i ) and ( ii ), we get

A = B = π/3
Then , C =π - π/3 - π/3 => C = π/3

Thius , sinA sinB sinC = [sin(π/3)]^3 = (3√3)/8

By taking some arbitrary values of A, B and C , We can check this
values is maximum or minimum

Suppose that A = π/2, B = π/3 and C = π/6

sinA sinB sinC = sin(π/2) sin(π/3) sin(π/6)
= 1(√3/2)(1/2)
= √3/4 < 3√3/8

Hence (3√3)/8 is the maximum value of sinA sinB
sinC

= sinA sinB sinC ≤ (3√3)/8

Option (A) is correct

Regards

Solve this: In a triangle ABC ( a ) sin A sin B sin C ≤ 3 3 8 ( b ) sin 2 A + sin 2 B + sin 2 C ≤ 9 4 ( c ) sin A sin B sin C i s a l w a y s p o s i t i v e ( d ) sin 2 A + sin 2 B ≤ 1 + cos C

Solve this:
In a triangle ABC
$(a) \sin A \sin B \sin C \leq \frac{3 \sqrt{3}}{8} (b) \sin^{2} A + \sin^{2} B + \sin^{2} C \leq \frac{9}{4} (c) \sin A \sin B \sin C i s a l w a y s p o s i t i v e (d) \sin^{2} A + \sin^{2} B \leq 1 + \cos C$