Question

In a triangle ABC ,find the maximum of sinAsinBsin(A+B) by using lagranges method
plZ Give the clear explanation of it ​

Answer 1

Step-by-step explanation:

*A2A*

sinx is concave in (0,π) , therefore,

sinA+sinB+sinC3≤sin(A+B+C3)(1)

Since, A, B, C are angles of a triangle, (1) gives,

sinA+sinB+sinC≤33–√2(2)

Since the upper bound in (2) can be easily achieved by setting A=B=C , the upper bound is tight. Hence the required maximum is 33√2 .

Answer 2

Step-by-step explanation:

fraction becomes 4/5 if 1 is added to both numerator and denominator.If however, 5 is

subtracted from both numerator and denominator the fraction becomes ½, what is the

fraction? Solve graphically

please graphically