Consider a triangle ABC with vertex A (2,-4). The internal bisectors of the angles B and C are x+y=2 and x-3 y=6 respectively. Let the two bisectors meet at I. Measure of the angle BIC equals

Question

Consider a triangle ABC with vertex A (2,-4). The internal bisectors of the angles B and C are x+y=2 and x-3 y=6 respectively. Let the two bisectors meet at I. Measure of the angle BIC equals (A) π- tan -1((1/2)) (B)  tan -1(2) (C) π- tan -1(2) (D) 2 tan -1(2)

Tardigrade Admin · Accepted Answer

text From figure, 135°=θ+θ1 text where tan θ1=(1/3) θ=135°-θ1 tan θ= tan (135°-θ1) =- tan (45°+θ1)=-((1+ tan θ1/1- tan θ1))--((4 / 3/2 / 3))=-2 ⇒ θ=π- tan -1 2

Q. Consider a triangle $A BC$ with vertex $A (2, - 4)$ . The internal bisectors of the angles $B$ and $C$ are $x + y = 2$ and $x - 3 y = 6$ respectively. Let the two bisectors meet at $I$ .
Measure of the angle BIC equals

$π - tan^{- 1} (\frac{1}{2})$

$tan^{- 1} (2)$

$π - tan^{- 1} (2)$

$2 tan^{- 1} (2)$

$From figure, 13 5^{\circ} = θ + θ_{1} where tan θ_{1} = \frac{1}{3}$
$θ = 13 5^{\circ} - θ_{1}$
$tan θ = tan (13 5^{\circ} - θ_{1})$
$= - tan (4 5^{\circ} + θ_{1}) = - (\frac{1 + t a n θ _{1}}{1 - t a n θ _{1}}) - - (\frac{4/3}{2/3}) = - 2$
$\Rightarrow θ = π - tan^{- 1} 2$

Q. Consider a triangle ABC with vertex A(2,−4). The internal bisectors of the angles B and C are x+y=2 and x−3y=6 respectively. Let the two bisectors meet at I. Measure of the angle BIC equals

π−tan−1(21​)

tan−1(2)

π−tan−1(2)

2tan−1(2)