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

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9.Straight Line

normal

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

Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $y -2 x =2$ such that $\triangle ABC$ is an equilateral triangle. Then, the area of the $\triangle ABC$ is

hard

(JEE MAIN-2023)

The line $\frac{x}{a} + \frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ where $a, b, c \in R_0$ and $c$ is constant, then locus of the foot of the perpendicular from the origin on the given line is –

normal

The triangle formed by the lines $x + y – 4 = 0,\,$ $3x + y = 4,$ $x + 3y = 4$ is

easy

(IIT-1983)

Let $A B C D$ be a square of side length $1$ . Let $P, Q, R, S$ be points in the interiors of the sides $A D, B C, A B, C D$ respectively, such that $P Q$ and $R S$ intersect at right angles. If $P Q=\frac{3 \sqrt{3}}{4}$, then $R S$ equals

normal

(KVPY-2015)

The equation of straight line passing through $( – a,\;0)$ and making the triangle with axes of area ‘$T$’ is

medium

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

Similar Questions

Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $y -2 x =2$ such that $\triangle ABC$ is an equilateral triangle. Then, the area of the $\triangle ABC$ is

The line $\frac{x}{a} + \frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ where $a, b, c \in R_0$ and $c$ is constant, then locus of the foot of the perpendicular from the origin on the given line is –

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