Question

$(K^4 +10K^2)/6=???$in triangle ABC cosAcosB+sinAsinBsinC=1 =>a:b:c=1:1:K then (K^4 +10K^2)/6=?

Answer 1

cosAcosB+sinAsinBsinC≤cosAcosB+sinAsinB

⇒1≤cos(A−B)⇒cos(A−B) = 1

⇒A= B and C= π/2

so,

a:b:c = 1 : 1 : √2

then value of K = √2

K⁴+10K²/6 = ?

K⁴ + 10K² /6

put value of K

→ (√2)⁴ + 10(√2)² /6

→ ( 4 + 10×2 )/6

→ ( 4+ 20 )/6

→ 24 /6

= 4

Answer 2

so,

a:b:c = 1 : 1 : √2

then value of K = √2

K⁴+10K²/6 = ?

K⁴ + 10K² /6

put value of K

(√2)⁴ + 10(√2)² /6

( 4 + 10×2 )/6

( 4+ 20 )/6

24 /6

= 4