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Question

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

$\frac{a d}{b c}$
$\frac{a b}{c d}$
$\frac{c d}{a b}$
$\frac{b c}{a d}$

The correct answer is: $fraction numerator b c over denominator a d end fraction$

Since a circle can be inscribed in the quadrilateral, thus $a plus c equals b plus d$
Further, it is also concyclic, thus $angle A plus angle C equals pi$
$B D to the power of 2 end exponent equals a to the power of 2 end exponent plus d to the power of 2 end exponent minus 2 a d cos invisible function application A$
Also, $blank B D to the power of 2 end exponent equals b to the power of 2 end exponent plus c to the power of 2 end exponent minus 2 b c cos invisible function application C equals b to the power of 2 end exponent plus c to the power of 2 end exponent plus 2 b c cos invisible function application A$
$rightwards double arrow 2 cos invisible function application A open parentheses b c plus a d close parentheses equals a to the power of 2 end exponent plus d to the power of 2 end exponent minus b to the power of 2 end exponent minus c to the power of 2 end exponent$

or $blank 2 cos invisible function application A equals fraction numerator a to the power of 2 end exponent plus d to the power of 2 end exponent minus b to the power of 2 end exponent minus c to the power of 2 end exponent over denominator b c plus a d end fraction$
Since $blank a plus c equals b plus d rightwards double arrow a minus d equals b minus c$
$rightwards double arrow a to the power of 2 end exponent plus d to the power of 2 end exponent minus 2 a d equals b to the power of 2 end exponent plus c to the power of 2 end exponent minus 2 b c$
$rightwards double arrow a to the power of 2 end exponent plus d to the power of 2 end exponent minus b to the power of 2 end exponent minus c to the power of 2 end exponent equals 2 open parentheses a d minus b c close parentheses$
$rightwards double arrow cos invisible function application A equals fraction numerator a d minus b c over denominator b c plus a d end fraction equals fraction numerator 1 minus tan to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction over denominator 1 plus tan to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction end fraction$
$rightwards double arrow tan to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction equals fraction numerator b c over denominator a d end fraction$

Related Questions to study

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

$log subscript 2 end subscript invisible function application x plus log subscript 2 end subscript invisible function application y greater or equal than 6 comma$ 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

parallel

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 $y element of open square brackets 0 comma blank 2 pi close square brackets right parenthesis$

$increment A B C comma$ 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

parallel

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

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

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

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

parallel

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 $y equals 2 plus sin invisible function application x a n d blank y equals cos invisible function application x blank i s$

e number of solutions of $2 sin to the power of 2 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application 2 x equals 2 comma blank x element of open square brackets 0 comma blank 2 pi close square brackets$ is

e number of solutions of $2 sin to the power of 2 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application 2 x equals 2 comma blank x element of open square brackets 0 comma blank 2 pi close square brackets$ is

parallel

e general solution of $tan invisible function application theta plus tan invisible function application 2 theta plus tan invisible function application 3 theta equals 0$ is

e general solution of $tan invisible function application theta plus tan invisible function application 2 theta plus tan invisible function application 3 theta equals 0$ is

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$alpha plus beta plus gamma equals 2 pi comma$ then

$alpha plus beta plus gamma equals 2 pi comma$ then

parallel

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