The angle bisectors BD and CE of a triangle A | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given,in $	riangle ABC$,the angle bisectors $BD$ and $CE$ are divided by the incentre $I$ in the ratios $3:2$ and $2:1$ respectively.$\frac{BI}{I

Vedclass · Accepted Answer

$11:4$

Vedclass · Answer

(B) Given,in $	riangle ABC$,the angle bisectors $BD$ and $CE$ are divided by the incentre $I$ in the ratios $3:2$ and $2:1$ respectively.$\frac{BI}{ID} = \frac{3}{2}$ and $\frac{CI}{IE} = \frac{2}{1}$.We know that for an incentre $I$ of $	riangle ABC$ with sides $a, b, c$ opposite to vertices $A, B, C$ respectively,the ratio in which $I$ divides the angle bisectors are:$\frac{AI}{IF} = \frac{b+c}{a}$,$\frac{BI}{ID} = \frac{a+c}{b}$,and $\frac{CI}{IE} = \frac{a+b}{c}$.Given $\frac{a+c}{b} = \frac{3}{2} \Rightarrow 2a + 2c = 3b \quad \dots(i)$Given $\frac{a+b}{c} = \frac{2}{1} \Rightarrow a + b = 2c \quad \dots(ii)$From $(ii)$,$c = \frac{a+b}{2}$. Substituting this into $(i)$:$2a + 2(\frac{a+b}{2}) = 3b$$2a + a + b = 3b$$3a = 2b \Rightarrow b = \frac{3}{2}a$Now,substitute $b = \frac{3}{2}a$ into $(ii)$:$a + \frac{3}{2}a = 2c$$\frac{5}{2}a = 2c \Rightarrow c = \frac{5}{4}a$Finally,the ratio $\frac{AI}{IF} = \frac{b+c}{a} = \frac{\frac{3}{2}a + \frac{5}{4}a}{a} = \frac{6+5}{4} = \frac{11}{4}$.Thus,the ratio is $11:4$.

The angle bisectors $BD$ and $CE$ of a $\triangle ABC$ are divided by the incentre $I$ in the ratios $3:2$ and $2:1$ respectively. Then,the ratio in which $I$ divides the angle bisector through $A$ is

Similar Questions

The equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$ is solvable for

Let $S = \{x \in [-\pi, \pi] : \sin x(\sin x + \cos x) = a, a \in Z\}$. Then $n(S)$ is equal to:

Let $\alpha$ and $\beta$ be two distinct roots of $a \cos \theta + b \sin \theta = c$, where $a, b, c$ are three real constants and $\theta \in [0, 2\pi]$. Then, $\alpha + \beta$ is also a root of the same equation, if

If $ABCD$ is a cyclic quadrilateral,then the value of $\cos A - \cos B + \cos C - \cos D = $

The equation $\sin 3\theta = 4 \sin \theta \sin 2\theta \sin 4\theta$ in the interval $0 \le \theta \le \pi$ has:

Vedclass Test Series

Exam Paper Generator

Online Exam Module