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Question

$increment A B C comma blank a comma blank b comma blank A$ 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

$(1 / 2) b^{2} \sin 2 A$
$(1 / 2) a^{2} \sin 2 A$
$b^{2} \sin 2 A$
None of these

The correct answer is: $open parentheses 1 divided by 2 close parentheses b to the power of 2 end exponent sin invisible function application 2 A$

We have $blank cos invisible function application A equals fraction numerator c to the power of 2 end exponent plus b to the power of 2 end exponent minus a to the power of 2 end exponent over denominator 2 blank b c end fraction rightwards double arrow c to the power of 2 end exponent minus 2 blank b c cos invisible function application A plus b to the power of 2 end exponent minus a to the power of 2 end exponent equals 0$
It is given that $c subscript 1 end subscript$ and $blank c subscript 2 end subscript$ are roots of this equation
Therefore, $c subscript 1 end subscript plus c subscript 2 end subscript equals 2 b cos invisible function application A blank$ and $blank c subscript 1 end subscript c subscript 2 end subscript equals b to the power of 2 end exponent minus a to the power of 2 end exponent$
$rightwards double arrow 2 R open parentheses sin invisible function application C subscript 1 end subscript plus sin invisible function application C subscript 2 end subscript close parentheses equals 4 blank R sin invisible function application B cos invisible function application A$
$rightwards double arrow sin invisible function application C subscript 1 end subscript plus sin invisible function application C subscript 2 end subscript equals 2 sin invisible function application B cos invisible function application A$
Now, sum of the areas of two triangles $equals fraction numerator 1 over denominator 2 end fraction blank a b sin invisible function application C subscript 1 end subscript plus fraction numerator 1 over denominator 2 end fraction blank a b sin invisible function application C subscript 2 end subscript$
$equals fraction numerator 1 over denominator 2 end fraction blank a b blank open parentheses sin invisible function application C subscript 1 end subscript plus sin invisible function application C subscript 2 end subscript close parentheses$
$equals fraction numerator 1 over denominator 2 end fraction blank a b blank open parentheses 2 sin invisible function application B cos invisible function application A close parentheses$
$equals a b sin invisible function application B cos invisible function application A equals b open parentheses b sin invisible function application A close parentheses cos invisible function application A equals fraction numerator 1 over denominator 2 end fraction blank b to the power of 2 end exponent sin invisible function application 2 A$

Related Questions to study

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

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

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

$sin invisible function application open parentheses alpha plus beta close parentheses equals 1 comma sin invisible function application open parentheses alpha minus beta close parentheses equals fraction numerator 1 over denominator 2 end fraction comma blank$ then $tan invisible function application open parentheses alpha plus 2 beta close parentheses tan invisible function application open parentheses 2 alpha plus beta close parentheses$ is equal to, $alpha comma blank beta element of open parentheses 0 comma blank pi divided by 2 close parentheses$

$sin invisible function application open parentheses alpha plus beta close parentheses equals 1 comma sin invisible function application open parentheses alpha minus beta close parentheses equals fraction numerator 1 over denominator 2 end fraction comma blank$ then $tan invisible function application open parentheses alpha plus 2 beta close parentheses tan invisible function application open parentheses 2 alpha plus beta close parentheses$ is equal to, $alpha comma blank beta element of open parentheses 0 comma blank pi divided by 2 close parentheses$

parallel

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

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

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

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

t area of triangle $A B C$ is $blank open parentheses square root of 3 minus 1 close parentheses divided by 2 comma blank b equals 2$ and $blank c equals open parentheses square root of 3 minus 1 close parentheses$ and $blank angle A$ is acute. The measure of the angle $C$ is

t area of triangle $A B C$ is $blank open parentheses square root of 3 minus 1 close parentheses divided by 2 comma blank b equals 2$ and $blank c equals open parentheses square root of 3 minus 1 close parentheses$ and $blank angle A$ is acute. The measure of the angle $C$ is

parallel

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