Question

In a triangle ABC if A = p/4 and tan B tan C = k, then k must satisfy –

• A. k² – 6k + 1 ³ 0
• B. k² – 6k + 1 = 0
• C. k² – 6k + 1 £ 0
• D. 3 – 2 $\sqrt { 2 }$ < k < 3 + 2$\sqrt { 2 }$

Answer 1

Answer: (A)

k² – 6k + 1 ³ 0

Answer 2

A = p/4, tan B tan C = k ̃ B + C = 3p/4

̃ tan (B + C) = tan $\frac { 3 \pi } { 4 }$

̃ $\frac { \tan B + \tan C } { 1 - \tan B \tan C }$ = –1 ̃ tan B + tan C = (k – 1)

Equation x² – (k – 1) x + k = 0

whole roots are tan B and tan C

for real vales D ³ 0 (k – 1)² – 4k ³ 0

̃ k² – 6k + 1 ³ 0.