In an isosceles triangle ABC, ∠ C = ∠ A if point of intersection of bisectors of inter

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In an isosceles triangle $ABC, \angle C = \angle A$ if point of intersection of bisectors of internal angles $\angle A$ and $\angle C$ divide median of side $AC$ in $3 : 1$ (from vertex $B$ to side $AC$), then value of $cosec \ \frac{B}{2}$ is equal to

$1$

$2$

$3$

$4$

Solution

As shown in above figure $\mathrm{AB}=\mathrm{BC}$

and $\mathrm{IB}=\mathrm{rcosec} \frac{\mathrm{B}}{2}, \mathrm{ID}=\mathrm{r}$

$ \Rightarrow \quad \frac{{{\rm{IB}}}}{{\rm{D}}} = \frac{{r\cos ec\frac{{\rm{B}}}{2}}}{{\rm{r}}} = \frac{3}{1}[{\rm{ where \,\,r\,\, is\,\, inradius }}]$

$ \Rightarrow \cos ec\frac{B}{2} = 3$

Standard 11

Mathematics

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In an isosceles triangle $ABC, \angle C = \angle A$ if point of intersection of bisectors of internal angles $\angle A$ and $\angle C$ divide median of side $AC$ in $3 : 1$ (from vertex $B$ to side $AC$), then value of $cosec \ \frac{B}{2}$ is equal to

Solution

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