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Question

$I$ is the incentre of a triangle $A B C comma$ then the ratio $I A colon I B colon I C$ is equal to

$c o s e c \frac{A}{2} : c o s e c \frac{B}{2} : c o s e c \frac{C}{2}$
$\sin \frac{A}{2} : \sin \frac{B}{2} : \sin \frac{C}{2}$
$\sec \frac{A}{2} : \sec \frac{B}{2} : \sec \frac{C}{2}$
None of these

The correct answer is: $c o s e c fraction numerator A over denominator 2 end fraction colon c o s e c fraction numerator B over denominator 2 end fraction colon c o s e c fraction numerator C over denominator 2 end fraction$

We know that $blank I A equals fraction numerator r over denominator sin invisible function application fraction numerator A over denominator 2 end fraction end fraction$
$rightwards double arrow I A colon I B colon I C equals c o s e c fraction numerator A over denominator 2 end fraction blank colon c o s e c fraction numerator B over denominator 2 end fraction blank colon c o s e c fraction numerator C over denominator 2 end fraction$

Related Questions to study

$log subscript 3 end subscript invisible function application left curly bracket 5 plus 4 log subscript 3 end subscript invisible function application open parentheses x minus 1 close parentheses right curly bracket blank equals 2 comma$ then $x$ is equal to

$log subscript 3 end subscript invisible function application left curly bracket 5 plus 4 log subscript 3 end subscript invisible function application open parentheses x minus 1 close parentheses right curly bracket blank equals 2 comma$ then $x$ is equal to

$log subscript y end subscript invisible function application x plus log subscript x end subscript invisible function application y equals 1 comma blank x to the power of 2 end exponent plus y equals 12 comma blank$ then the value of $x y$ is

$log subscript y end subscript invisible function application x plus log subscript x end subscript invisible function application y equals 1 comma blank x to the power of 2 end exponent plus y equals 12 comma blank$ then the value of $x y$ is

e number of solutions of the equation $tan invisible function application x plus sec invisible function application x minus 2 cos invisible function application x$ lying in the interval $left square bracket 0 comma blank 2 pi right square bracket$ is

e number of solutions of the equation $tan invisible function application x plus sec invisible function application x minus 2 cos invisible function application x$ lying in the interval $left square bracket 0 comma blank 2 pi right square bracket$ is

parallel

a convex quadrilateral $A B C D comma blank A B equals a comma blank B C equals b comma blank C D equals c blank$ and $D A equals d.$ This quadrilateral is such that a circle can be inscribed in it and a circle can be also circumscribed about it, then $tan to the power of 2 end exponent invisible function application open parentheses A divided by 2 close parentheses$ is equal to

a convex quadrilateral $A B C D comma blank A B equals a comma blank B C equals b comma blank C D equals c blank$ and $D A equals d.$ This quadrilateral is such that a circle can be inscribed in it and a circle can be also circumscribed about it, then $tan to the power of 2 end exponent invisible function application open parentheses A divided by 2 close parentheses$ is equal to

triangle $A B C comma$ line joining the circumcentre and orthocenter is parallel to side $A C comma$ then the value of $tan invisible function application A tan invisible function application C$ is equal to

triangle $A B C comma$ line joining the circumcentre and orthocenter is parallel to side $A C comma$ then the value of $tan invisible function application A tan invisible function application C$ is equal to

$increment A B C comma blank a comma blank b comma blank A$ are given and $c subscript 1 end subscript comma c subscript 2 end subscript$ are two values of the third side $c.$ The sum of the areas of the two triangles with sides $a comma blank b comma blank c subscript 1 end subscript$ and $blank a comma blank b comma blank c subscript 2 end subscript$ is

$increment A B C comma blank a comma blank b comma blank A$ are given and $c subscript 1 end subscript comma c subscript 2 end subscript$ are two values of the third side $c.$ The sum of the areas of the two triangles with sides $a comma blank b comma blank c subscript 1 end subscript$ and $blank a comma blank b comma blank c subscript 2 end subscript$ is

parallel

$log subscript 2 end subscript invisible function application x plus log subscript 2 end subscript invisible function application y greater or equal than 6 comma$ then the least value of $x plus y$ is

$log subscript 2 end subscript invisible function application x plus log subscript 2 end subscript invisible function application y greater or equal than 6 comma$ then the least value of $x plus y$ is

t $a greater than 1$ be a real number. Then the number of roots equation $a to the power of 2 log subscript 2 end subscript invisible function application x end exponent equals 5 plus 4 x to the power of log subscript 2 end subscript invisible function application a end exponent$ has

t $a greater than 1$ be a real number. Then the number of roots equation $a to the power of 2 log subscript 2 end subscript invisible function application x end exponent equals 5 plus 4 x to the power of log subscript 2 end subscript invisible function application a end exponent$ has

of ordered pairs which satisfy the equation $x to the power of 2 end exponent plus 2 x sin invisible function application open parentheses x y close parentheses plus 1 equals 0$ are $blank$ (where $y element of open square brackets 0 comma blank 2 pi close square brackets right parenthesis$

of ordered pairs which satisfy the equation $x to the power of 2 end exponent plus 2 x sin invisible function application open parentheses x y close parentheses plus 1 equals 0$ are $blank$ (where $y element of open square brackets 0 comma blank 2 pi close square brackets right parenthesis$

parallel

$increment A B C comma$ if $sin to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction comma sin to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction$ and $sin to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction$ are in H.P., then $blank a comma blank b blank$ and $blank c blank$ will be in

$increment A B C comma$ if $sin to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction comma sin to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction$ and $sin to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction$ are in H.P., then $blank a comma blank b blank$ and $blank c blank$ will be in

Member of solutions of $tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction sin invisible function application theta close parentheses equals cot invisible function application open parentheses fraction numerator pi over denominator 2 end fraction cos invisible function application theta close parentheses comma blank theta element of open square brackets 0 comma blank 6 pi close square brackets comma blank$ is

Member of solutions of $tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction sin invisible function application theta close parentheses equals cot invisible function application open parentheses fraction numerator pi over denominator 2 end fraction cos invisible function application theta close parentheses comma blank theta element of open square brackets 0 comma blank 6 pi close square brackets comma blank$ is

$i s blank e q u a l blank t o$

$i s blank e q u a l blank t o$

parallel

$sin to the power of 2 end exponent invisible function application theta equals fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent plus 1 over denominator 2 x end fraction comma blank$ then $x blank$ must be

$sin to the power of 2 end exponent invisible function application theta equals fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent plus 1 over denominator 2 x end fraction comma blank$ then $x blank$ must be

e equation $cos to the power of 8 end exponent invisible function application x plus b cos to the power of 4 end exponent invisible function application x plus 1 equals 0$ will have a solution if $b$ belongs to

e equation $cos to the power of 8 end exponent invisible function application x plus b cos to the power of 4 end exponent invisible function application x plus 1 equals 0$ will have a solution if $b$ belongs to

e minimum vertical distance between the graphs of $y equals 2 plus sin invisible function application x a n d blank y equals cos invisible function application x blank i s$

e minimum vertical distance between the graphs of $y equals 2 plus sin invisible function application x a n d blank y equals cos invisible function application x blank i s$

parallel

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