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Maths-

General

Easy

Question

The radius of the circle passing through the centre or the in-circle of $triangle A B C$ and through the end points of BC is given by

$\frac{a}{2} c o s A$
$\frac{a}{2} s e c A / 2$
$\frac{a}{2} s i n A$
$a s e c A / 2$

The correct answer is: $fraction numerator a over denominator 2 end fraction s e c invisible function application A divided by 2$

$because angle B I C equals 180 to the power of ring operator end exponent minus open parentheses fraction numerator B over denominator 2 end fraction plus fraction numerator C over denominator 2 end fraction close parentheses$
$S o angle B O C equals 2 open parentheses 180 to the power of ring operator end exponent minus open parentheses fraction numerator B over denominator 2 end fraction plus fraction numerator C over denominator 2 end fraction close parentheses close parentheses$
$equals 360 to the power of ring operator end exponent minus left parenthesis B plus C right parenthesis$
$open curly brackets therefore A plus B plus C equals 180 to the power of ring operator end exponent close parentheses$
$equals pi minus A$
$l n invisible function application capital delta O B C$
$c o s invisible function application left parenthesis pi minus A right parenthesis equals fraction numerator R to the power of 2 end exponent plus R to the power of 2 end exponent minus a to the power of 2 end exponent over denominator 2. R R R end fraction$
$rightwards double arrow a to the power of 2 end exponent equals 2 R to the power of 2 end exponent left parenthesis 1 plus c o s invisible function application A right parenthesis$
$R equals fraction numerator a over denominator square root of 4 c o s to the power of 2 end exponent invisible function application fraction numerator 2 A over denominator 2 end fraction end root end fraction equals fraction numerator a over denominator 2 end fraction s e c invisible function application fraction numerator A over denominator 2 end fraction$

Related Questions to study

CH₃COOH is reacted with $H C identical to C H$ in presence of Hg²⁺ , the product is:

CH₃COOH is reacted with $H C identical to C H$ in presence of Hg²⁺ , the product is:

Chemistry-General

If $fraction numerator 1 over denominator a end fraction comma fraction numerator 1 over denominator b end fraction comma fraction numerator 1 over denominator c end fraction$ are in A.P., where a, b, c are sides of the triangle ABC, then $s i n to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction comma s i n to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction comma s i n to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction$ are in

If $fraction numerator 1 over denominator a end fraction comma fraction numerator 1 over denominator b end fraction comma fraction numerator 1 over denominator c end fraction$ are in A.P., where a, b, c are sides of the triangle ABC, then $s i n to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction comma s i n to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction comma s i n to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction$ are in

Statement1: If a line intersects a hyperbola at (2, 6) and (4, 4) and one of the asymptotes at (1, 2) then the centre of the hyperbola is (1, 2) because
Statement2: Mid point of the chord intercepted by hyperbola is same as midpoint of the chord intercepted between
asymptotes

Statement1: If a line intersects a hyperbola at (2, 6) and (4, 4) and one of the asymptotes at (1, 2) then the centre of the hyperbola is (1, 2) because
Statement2: Mid point of the chord intercepted by hyperbola is same as midpoint of the chord intercepted between
asymptotes

parallel

lf $not stretchy integral subscript blank superscript blank left parenthesis blank s i n blank 2 x plus blank c o s blank 2 x right parenthesis d x equals fraction numerator 1 over denominator square root of 2 end fraction blank s i n blank left parenthesis 2 x minus c right parenthesis plus 0$ , then the value of a and c is

lf $not stretchy integral subscript blank superscript blank left parenthesis blank s i n blank 2 x plus blank c o s blank 2 x right parenthesis d x equals fraction numerator 1 over denominator square root of 2 end fraction blank s i n blank left parenthesis 2 x minus c right parenthesis plus 0$ , then the value of a and c is

lf $integral subscript blank superscript blank fraction numerator left parenthesis x squared minus 1 right parenthesis over denominator left parenthesis x to the power of 4 plus 3 x squared plus 1 right parenthesis t a n to the power of negative 1 end exponent left parenthesis fraction numerator x squared plus right square bracket over denominator chi end fraction right parenthesis end fraction d x equals k blank l o g blank vertical line t a n to the power of negative 1 end exponent fraction numerator x squared plus 1 over denominator chi end fraction vertical line plus c comma$ then $k$ is equal to

lf $integral subscript blank superscript blank fraction numerator left parenthesis x squared minus 1 right parenthesis over denominator left parenthesis x to the power of 4 plus 3 x squared plus 1 right parenthesis t a n to the power of negative 1 end exponent left parenthesis fraction numerator x squared plus right square bracket over denominator chi end fraction right parenthesis end fraction d x equals k blank l o g blank vertical line t a n to the power of negative 1 end exponent fraction numerator x squared plus 1 over denominator chi end fraction vertical line plus c comma$ then $k$ is equal to

lf $integral subscript blank superscript blank fraction numerator d x over denominator chi square root of 1 minus x cubed end root end fraction equals a blank l o g blank vertical line fraction numerator square root of 1 minus x cubed end root minus 1 over denominator square root of 1 minus x cubed end root plus 1 end fraction vertical line plus c$ , then

lf $integral subscript blank superscript blank fraction numerator d x over denominator chi square root of 1 minus x cubed end root end fraction equals a blank l o g blank vertical line fraction numerator square root of 1 minus x cubed end root minus 1 over denominator square root of 1 minus x cubed end root plus 1 end fraction vertical line plus c$ , then

parallel

The gradient of the curve y=f(x) at P is slope of the tangent line at P i.e.,

PM = length of the tangent
PR = length of the normal to the curve y=f(x)
Area of triangle formed by the positive x-axis and the normal and the tangent to $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 4$ at $left parenthesis 1 comma square root of 3 right parenthesis$ is

The gradient of the curve y=f(x) at P is slope of the tangent line at P i.e.,

PM = length of the tangent
PR = length of the normal to the curve y=f(x)
Area of triangle formed by the positive x-axis and the normal and the tangent to $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 4$ at $left parenthesis 1 comma square root of 3 right parenthesis$ is

The partial pressures of CH₃0H, CO and H₂ in the equilibrium mixture for the reaction, $C O plus 2 H subscript 2 end subscript rightwards harpoon over leftwards harpoon C H subscript 3 end subscript O H$ at 427 $degree$ C are 2.0, 1.0 and 0.1 atm respectively The value of K_p for the decomposition of CH₃0H into CO and H₂ is:

The partial pressures of CH₃0H, CO and H₂ in the equilibrium mixture for the reaction, $C O plus 2 H subscript 2 end subscript rightwards harpoon over leftwards harpoon C H subscript 3 end subscript O H$ at 427 $degree$ C are 2.0, 1.0 and 0.1 atm respectively The value of K_p for the decomposition of CH₃0H into CO and H₂ is:

Chemistry-General

At 500 K, the equilibrium constant for reaction c is $C subscript 2 end subscript H subscript 2 end subscript C l subscript 2 end subscript rightwards harpoon over leftwards harpoon text trans- end text C subscript 2 end subscript H subscript 2 end subscript C l subscript 2 end subscript$ 0.6At the same temperature, the equilibrium constant for the reaction $C subscript 2 end subscript H subscript 2 end subscript C l subscript 2 end subscript rightwards harpoon over leftwards harpoon text trans- end text C subscript 2 end subscript H subscript 2 end subscript C l subscript 2 end subscript$ , will be:

At 500 K, the equilibrium constant for reaction c is $C subscript 2 end subscript H subscript 2 end subscript C l subscript 2 end subscript rightwards harpoon over leftwards harpoon text trans- end text C subscript 2 end subscript H subscript 2 end subscript C l subscript 2 end subscript$ 0.6At the same temperature, the equilibrium constant for the reaction $C subscript 2 end subscript H subscript 2 end subscript C l subscript 2 end subscript rightwards harpoon over leftwards harpoon text trans- end text C subscript 2 end subscript H subscript 2 end subscript C l subscript 2 end subscript$ , will be:

Chemistry-General

parallel

For the reaction, $blank 2 N O subscript 2 end subscript left parenthesis g right parenthesis rightwards harpoon over leftwards harpoon 2 N O left parenthesis g right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis$ $K subscript c end subscript equals 1.8 cross times 10 to the power of negative 6 end exponent$ KC 1.8 $cross times$ 10^-6 at 185 $degree$ C The value of Kc at 185 $degree$ C for the reaction; $N O subscript 2 end subscript left parenthesis g right parenthesis rightwards harpoon over leftwards harpoon fraction numerator 1 over denominator 2 end fraction N subscript 2 end subscript left parenthesis g right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis$

For the reaction, $blank 2 N O subscript 2 end subscript left parenthesis g right parenthesis rightwards harpoon over leftwards harpoon 2 N O left parenthesis g right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis$ $K subscript c end subscript equals 1.8 cross times 10 to the power of negative 6 end exponent$ KC 1.8 $cross times$ 10^-6 at 185 $degree$ C The value of Kc at 185 $degree$ C for the reaction; $N O subscript 2 end subscript left parenthesis g right parenthesis rightwards harpoon over leftwards harpoon fraction numerator 1 over denominator 2 end fraction N subscript 2 end subscript left parenthesis g right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis$

Chemistry-General

In a geometric progression, if the ratio of the sum of first 5 terms to the sum of their reciprocals is 49, and the sum of the first and third terms is 35, then the first term of this geometric progression is

In a geometric progression, if the ratio of the sum of first 5 terms to the sum of their reciprocals is 49, and the sum of the first and third terms is 35, then the first term of this geometric progression is

Statements-1: $stretchy integral subscript 0 end subscript superscript n x plus t end superscript vertical line s i n invisible function application x vertical line d x$ = (2n + (a) cost (0 t $theta$ )
Statements-2: $stretchy integral subscript a end subscript superscript b end superscript f left parenthesis x right parenthesis d x equals stretchy integral subscript a end subscript superscript c end superscript f left parenthesis x right parenthesis d x plus stretchy integral subscript c end subscript superscript b end superscript f left parenthesis x right parenthesis d x$
and $stretchy integral subscript 0 end subscript superscript m end superscript f left parenthesis x right parenthesis d x equals n stretchy integral subscript 0 end subscript superscript a end superscript f left parenthesis x right parenthesis d x$ if f(a + x) = f(x)

Statements-1: $stretchy integral subscript 0 end subscript superscript n x plus t end superscript vertical line s i n invisible function application x vertical line d x$ = (2n + (a) cost (0 t $theta$ )
Statements-2: $stretchy integral subscript a end subscript superscript b end superscript f left parenthesis x right parenthesis d x equals stretchy integral subscript a end subscript superscript c end superscript f left parenthesis x right parenthesis d x plus stretchy integral subscript c end subscript superscript b end superscript f left parenthesis x right parenthesis d x$
and $stretchy integral subscript 0 end subscript superscript m end superscript f left parenthesis x right parenthesis d x equals n stretchy integral subscript 0 end subscript superscript a end superscript f left parenthesis x right parenthesis d x$ if f(a + x) = f(x)

parallel

If x enotes the rounded off value of x $left parenthesis 3.4 equals 3 comma 3.5 equals 4 comma 3.8 equals text 4etc end text right parenthesis$ and the area bounded by the two curves defined by y = $vertical line x minus x with not stretchy bar on top vertical line$ and $y equals lambda sin invisible function application left parenthesis pi x right parenthesis comma lambda 1$ . In the interval 0 x 1 is equal to 2 square units, The value of the constant is

If x enotes the rounded off value of x $left parenthesis 3.4 equals 3 comma 3.5 equals 4 comma 3.8 equals text 4etc end text right parenthesis$ and the area bounded by the two curves defined by y = $vertical line x minus x with not stretchy bar on top vertical line$ and $y equals lambda sin invisible function application left parenthesis pi x right parenthesis comma lambda 1$ . In the interval 0 x 1 is equal to 2 square units, The value of the constant is

Statement-1: $f colon R rightwards arrow R comma f left parenthesis x right parenthesis equals s i n invisible function application x plus x$ then $stretchy integral subscript 0 end subscript superscript pi end superscript open parentheses f to the power of negative 1 end exponent left parenthesis x right parenthesis close parentheses d x equals fraction numerator pi to the power of 2 end exponent minus 4 over denominator 2 end fraction$
Statement-2 : $y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis$ is the reflection of $y equals f left parenthesis x right parenthesis text w.r.t. end text to the power of ´ end exponent x to the power of ´ end exponent$

Statement-1: $f colon R rightwards arrow R comma f left parenthesis x right parenthesis equals s i n invisible function application x plus x$ then $stretchy integral subscript 0 end subscript superscript pi end superscript open parentheses f to the power of negative 1 end exponent left parenthesis x right parenthesis close parentheses d x equals fraction numerator pi to the power of 2 end exponent minus 4 over denominator 2 end fraction$
Statement-2 : $y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis$ is the reflection of $y equals f left parenthesis x right parenthesis text w.r.t. end text to the power of ´ end exponent x to the power of ´ end exponent$

If $stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent invisible function application x d x equals fraction numerator pi over denominator 4 end fraction minus fraction numerator 1 over denominator 2 end fraction$ ln 2 then the value of the definite integral $stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent invisible function application open parentheses 1 minus x plus x to the power of 2 end exponent close parentheses d x$ equals

If $stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent invisible function application x d x equals fraction numerator pi over denominator 4 end fraction minus fraction numerator 1 over denominator 2 end fraction$ ln 2 then the value of the definite integral $stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent invisible function application open parentheses 1 minus x plus x to the power of 2 end exponent close parentheses d x$ equals

parallel

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