Question

Answer.................​

Answer 1

✪ANSWER✪

• The least positive integral value of K is 6.

★EXPLAINATION★

☆GIVEN☆

• Triangle ABC such that $A = \frac{π}{4}$

• $tan(B)tan(C) = K.....(1)$

☆TO FIND☆

• Least positive integral value of K.

☆CALCULATION☆

Now,

$A+B+C = π$

$\frac{π}{4} +B+C = π$

$B+C = π - \frac{π}{4}$

$B+C = \frac{3π}{4}$

$tan(B+C) = tan(\frac{3π}{4} )$

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

$tan(B)+tan(C) = -1 + tan(B)tan(C)$

$tan(B)+tan(C) = (K-1).....(2)$

We know that tan(B) and tan(C) are roots of

$x² -[tan(B) + tan(C)]x + tan(B)tan(C) = 0$

Using (1) and (2)

$x² -[K-1]x + K = 0$

Since tan(B) and tan(C) are real,

Discriminant ≥ 0

$⇒ (K-1)² - 4K ≥ 0$

$⇒ K² -2K + 1 - 4K ≥ 0$

$⇒ K² -6K + 1≥ 0$

$⇒ K² -2K.3 +3²+ 1-3²≥ 0$

$⇒ (K -3)²- (2√2)²≥ 0$

$⇒ [K-3+2√2][K -3-2√2)1] ≥ 0$

$⇒ [K-(3-2√2)][K -(3+2√2)1] ≥ 0$

$⇒ K ≤ (3-2√2)\:\:or\:\:K ≥ (3+2√2)$

Therefore least positive integral value of K

= [3+ 2√2] + 1, [.] is greatest integer function.

= 5 + 1

= 6