Answer :
|D
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A| | 2x | |
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x | |
| | C |................k/2................B................k/2...............| E
<---------------------------------k-------------------------->
join the point A to B and B to D to make Δ
let <ABC = α and <DBE = 90 - α
in ΔABC
tanα = AC/BC = x/(k/2)
tanα = 2x/k ----------------(1)
in ΔDBE
tan(90-α) = DE/BE = 2x/(k/2)
cotα = 4x/k
1/tanα = 4x/k
put the value of tanα from the (1) equation
1/(2x/k) = 4x/k
k/2x = 4x/k
8x² = k²
x² = k²/8
x = k/2√2 meter
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|
|
|
A| | 2x | |
| |
x | |
| | C |................k/2................B................k/2...............| E
<---------------------------------k-------------------------->
join the point A to B and B to D to make Δ
let <ABC = α and <DBE = 90 - α
in ΔABC
tanα = AC/BC = x/(k/2)
tanα = 2x/k ----------------(1)
in ΔDBE
tan(90-α) = DE/BE = 2x/(k/2)
cotα = 4x/k
1/tanα = 4x/k
put the value of tanα from the (1) equation
1/(2x/k) = 4x/k
k/2x = 4x/k
8x² = k²
x² = k²/8
x = k/2√2 meter
dc=2*ab
ae=ec
angle e=x
then angle aeb= 90-x
so angle aeb= 1/tanx
look into the triangle edc
tan x=[tex] \frac{dc}{ec} [/tex]
=[tex] \frac{2ab}{ae} [/tex]...(1)
look into the triangle bae
tan [tex] \frac{1}{tan/x}= \frac{ab}{ae} \\ \\ tan/x*ab=ae[/tex]
[tex]tanx= \frac{ae}{ab} [/tex]...(2)
1=2
[tex] \frac{ae}{ab} = \frac{2ab}{ae} \\ \\ ae ^{2} =2a b ^{2} \\ ae=k/2 \\ ae ^{2} = \frac{k ^{2} }{4} \\ \\ \frac{k ^{2} }{4} =2ab ^{2} \\ ab ^{2} = \frac{k ^{2} }{8} \\ ab= \sqrt{ \frac{k ^{2} }{8} }[/tex]
ae=ec
angle e=x
then angle aeb= 90-x
so angle aeb= 1/tanx
look into the triangle edc
tan x=[tex] \frac{dc}{ec} [/tex]
=[tex] \frac{2ab}{ae} [/tex]...(1)
look into the triangle bae
tan [tex] \frac{1}{tan/x}= \frac{ab}{ae} \\ \\ tan/x*ab=ae[/tex]
[tex]tanx= \frac{ae}{ab} [/tex]...(2)
1=2
[tex] \frac{ae}{ab} = \frac{2ab}{ae} \\ \\ ae ^{2} =2a b ^{2} \\ ae=k/2 \\ ae ^{2} = \frac{k ^{2} }{4} \\ \\ \frac{k ^{2} }{4} =2ab ^{2} \\ ab ^{2} = \frac{k ^{2} }{8} \\ ab= \sqrt{ \frac{k ^{2} }{8} }[/tex]
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