plz answer question no. 11
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Let height of tower be 'h'.
Let the distance CB be 'x'.
In ∆ABC,
AB/BC = tan60°
h/x = √3
h = √3x
In ∆ABD ,
AB/BD = tan30°
h/x+20 = 1/√3
20+x = h√3 {putting value of h}
20+x = √3x × √3
20+x = 3x
20 = 3x-x
20 = 2x
x= 10
Total length of canal is 30m.
Height of tower is 10√3.
Let the distance CB be 'x'.
In ∆ABC,
AB/BC = tan60°
h/x = √3
h = √3x
In ∆ABD ,
AB/BD = tan30°
h/x+20 = 1/√3
20+x = h√3 {putting value of h}
20+x = √3x × √3
20+x = 3x
20 = 3x-x
20 = 2x
x= 10
Total length of canal is 30m.
Height of tower is 10√3.
Jasmattu:
Perfect answer
thnx
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