In /_\abc if ad is bisector of bac and de bisects [8da,b=40^(@)-Turito

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Maths-

In $triangle A B C$ if AD is bisector of $stack B A C with blank below$ and DE bisects $open floor B D A comma open floor B equals 40 to the power of ring operator end exponent comma open floor C equals 80 to the power of ring operator end exponent close close close$ find $left floor B E D.$

Maths-General

$85^{0}$
$95^{0}$
$75^{0}$
$105^{0}$

Answer:The correct answer is: $105 to the power of 0 end exponent$

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Related Questions to study

In given figure $stack A B with rightwards arrow on top equals stack A C with bar on top$ and $stack B C with bar on top equals stack C D with bar on top comma$ if $B A C equals 80 to the power of ring operator end exponent blank$ then find $∣ C D B$

In given figure $stack A B with rightwards arrow on top equals stack A C with bar on top$ and $stack B C with bar on top equals stack C D with bar on top comma$ if $B A C equals 80 to the power of ring operator end exponent blank$ then find $∣ C D B$

In the diagram, if $triangle A B C$ and $capital delta P Q R$ are Equilateral then find $text C end text X Y$

In the diagram, if $triangle A B C$ and $capital delta P Q R$ are Equilateral then find $text C end text X Y$

if $a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent equals a b plus b c plus c a$ then the triangle is

if $a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent equals a b plus b c plus c a$ then the triangle is

A train moves from one station to another 2 hours time. Its speed-time graph during this motion is shown in the figure. The maximum acceleration during the journey is

Maximum acceleration will be represented by

C D

part of the graph
Acceleration

equals fraction numerator d v over denominator d t end fraction equals fraction numerator left parenthesis 60 minus 20 right parenthesis over denominator 0.25 end fraction equals 160 blank k m divided by h to the power of 2 end exponent

A train moves from one station to another 2 hours time. Its speed-time graph during this motion is shown in the figure. The maximum acceleration during the journey is

physics-General

Maximum acceleration will be represented by

C D

part of the graph
Acceleration

equals fraction numerator d v over denominator d t end fraction equals fraction numerator left parenthesis 60 minus 20 right parenthesis over denominator 0.25 end fraction equals 160 blank k m divided by h to the power of 2 end exponent

The position of a particle at any instant $t$ is given by $x equals a$ cos $omega t$ . The speed-time graph of the particle is

therefore blank x equals a

omega t

therefore v equals fraction numerator d x over denominator d t end fraction equals negative a omega sin invisible function application omega t

The instantaneous speed is given by modulus of instantaneous velocity.

therefore blank

equals open vertical bar u close vertical bar equals open vertical bar negative a omega sin invisible function application omega t close vertical bar

Hence, [c] is correct.

The position of a particle at any instant $t$ is given by $x equals a$ cos $omega t$ . The speed-time graph of the particle is

physics-General

therefore blank x equals a

omega t

therefore v equals fraction numerator d x over denominator d t end fraction equals negative a omega sin invisible function application omega t

The instantaneous speed is given by modulus of instantaneous velocity.

therefore blank

equals open vertical bar u close vertical bar equals open vertical bar negative a omega sin invisible function application omega t close vertical bar

Hence, [c] is correct.

The velocity-time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in $6 blank s e c$ are respectively

Displacement = Summation of all the area with sign

equals open parentheses A subscript 1 end subscript close parentheses plus open parentheses negative A subscript 2 end subscript close parentheses plus open parentheses A subscript 3 end subscript close parentheses equals open parentheses 2 cross times 4 close parentheses plus open parentheses negative 2 cross times 2 close parentheses plus open parentheses 2 cross times 2 close parentheses

therefore D i s p l a c e m e n t equals 8 blank m

Distance = Summation of all the areas without sign

equals open vertical bar A subscript 1 end subscript close vertical bar plus open vertical bar negative A subscript 2 end subscript close vertical bar plus open vertical bar A subscript 3 end subscript close vertical bar equals open vertical bar 8 close vertical bar plus open vertical bar negative 4 close vertical bar plus open vertical bar 4 close vertical bar equals 8 plus 4 plus 4

therefore D i s t a n c e equals 16 blank m

The velocity-time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in $6 blank s e c$ are respectively

physics-General

Displacement = Summation of all the area with sign

equals open parentheses A subscript 1 end subscript close parentheses plus open parentheses negative A subscript 2 end subscript close parentheses plus open parentheses A subscript 3 end subscript close parentheses equals open parentheses 2 cross times 4 close parentheses plus open parentheses negative 2 cross times 2 close parentheses plus open parentheses 2 cross times 2 close parentheses

therefore D i s p l a c e m e n t equals 8 blank m

Distance = Summation of all the areas without sign

equals open vertical bar A subscript 1 end subscript close vertical bar plus open vertical bar negative A subscript 2 end subscript close vertical bar plus open vertical bar A subscript 3 end subscript close vertical bar equals open vertical bar 8 close vertical bar plus open vertical bar negative 4 close vertical bar plus open vertical bar 4 close vertical bar equals 8 plus 4 plus 4

therefore D i s t a n c e equals 16 blank m

$triangle A B C$ is an Isosceles Triangle, if $A equals 92 to the power of ring operator end exponent$ then find ABD .

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Lines PS, QT and RU intersect at a common point ‘O’. P is joined to Q, R to S and T to U to form triangles. Then find the value of $angle P plus angle Q plus angle R plus angle S plus angle T plus angle U equals$

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find the value of x in given figure. $blank l parallel to m$ and $text PI|Q end text$

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In the given figure, $angle 1 equals 8 x minus 6 to the power of ring operator end exponent comma angle 3 equals 3 x plus 4 to the power of ring operator end exponent$ and then find $angle 4 equals 4 x plus 2 to the power of ring operator end exponent$ then find $angle 1.$

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The value of $not stretchy integral subscript 0 end subscript superscript pi end superscript open vertical bar sin to the power of 3 end exponent invisible function application rightwards arrow over short leftwards arrow theta close vertical bar d theta$ is

I equals not stretchy integral subscript 0 end subscript superscript pi end superscript vertical line sin to the power of 3 end exponent invisible function application theta vertical line d theta

sin invisible function application theta

is positive in interval

left parenthesis 0 comma pi right parenthesis

therefore I equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin to the power of 3 end exponent invisible function application theta d theta equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin invisible function application theta left parenthesis 1 minus cos to the power of 2 end exponent invisible function application theta right parenthesis d theta

equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin invisible function application theta d theta plus not stretchy integral subscript 0 end subscript superscript pi end superscript left parenthesis negative sin invisible function application theta right parenthesis cos to the power of 2 end exponent invisible function application theta d theta

equals left square bracket negative cos invisible function application theta right square bracket subscript 0 end subscript superscript pi end superscript plus open parentheses fraction numerator cos to the power of 3 end exponent invisible function application theta over denominator 3 end fraction close parentheses subscript 0 end subscript superscript pi end superscript equals fraction numerator 4 over denominator 3 end fraction

The value of $not stretchy integral subscript 0 end subscript superscript pi end superscript open vertical bar sin to the power of 3 end exponent invisible function application rightwards arrow over short leftwards arrow theta close vertical bar d theta$ is

I equals not stretchy integral subscript 0 end subscript superscript pi end superscript vertical line sin to the power of 3 end exponent invisible function application theta vertical line d theta

sin invisible function application theta

is positive in interval

left parenthesis 0 comma pi right parenthesis

therefore I equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin to the power of 3 end exponent invisible function application theta d theta equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin invisible function application theta left parenthesis 1 minus cos to the power of 2 end exponent invisible function application theta right parenthesis d theta

equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin invisible function application theta d theta plus not stretchy integral subscript 0 end subscript superscript pi end superscript left parenthesis negative sin invisible function application theta right parenthesis cos to the power of 2 end exponent invisible function application theta d theta

equals left square bracket negative cos invisible function application theta right square bracket subscript 0 end subscript superscript pi end superscript plus open parentheses fraction numerator cos to the power of 3 end exponent invisible function application theta over denominator 3 end fraction close parentheses subscript 0 end subscript superscript pi end superscript equals fraction numerator 4 over denominator 3 end fraction

A spherical shell, made of material of electrical conductivity $fraction numerator 10 to the power of 9 end exponent over denominator pi end fraction left parenthesis capital omega minus m right parenthesis to the power of negative 1 end exponent$ has + - thickness t = 2 mm and radius r = 10 cm. In an arrangement, its inside surface is kept at a lower potential than its outside surface The resistance offered by the shell is equal to
"

A spherical shell, made of material of electrical conductivity $fraction numerator 10 to the power of 9 end exponent over denominator pi end fraction left parenthesis capital omega minus m right parenthesis to the power of negative 1 end exponent$ has + - thickness t = 2 mm and radius r = 10 cm. In an arrangement, its inside surface is kept at a lower potential than its outside surface The resistance offered by the shell is equal to
"

physics-General

In the shown wire frame, each side of a square (the smallest square) has a resistance R. The equivalent resistance of the circuit between the points A and B is :

In the shown wire frame, each side of a square (the smallest square) has a resistance R. The equivalent resistance of the circuit between the points A and B is :

physics-General