Latest CAT Aptitude Question SOLUTION: On the level ground, the angle of elevation of the top of the tower is 30degrees, on moving 20 metres nearer; the angle of elevation is 60degrees. Then the height of the tower is

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On the level ground, the angle of elevation of the top of the tower is 30degrees, on moving 20 metres nearer; the angle of elevation is 60degrees. Then the height of the tower is

Asked In CAT chaitanya (11 years ago)

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kiran ks  (11 years ago)

First draw the right angle triangles according to the given condition.
From rt angle triangle,
tan60=h/x (h=height of the tower and x= dist between bottom of tower to 2nd
root3=h/x point where the angle of elevation is 60 degree)
also, tan30=h/(20+x)
1/root3=h/(20+x)
From above two equations, x*root3=(20+x)/root3
3x=20+x
so, x=10
Then, h=x*root3=17.32m

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prashanthi  (11 years ago)

draw triangle BCA BCIS HYPOTHESIS BA AS PERPENDICULAR AND CA AS BASE
NOW AB/AD=tan60
AD=AB/ROOT3=h/ROOT3
AB/AC=tan30
AC=AB*ROOT3=h*ROOT3
CD=(AC-AD)
(h*ROOT3-h/ROOT3)=24
h=12*root3
h=20.76m

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sonam tripathy  (11 years ago)

draw a rt triangle ABC in which

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vishnu  (11 years ago)

First of all draw a right angle triangle ABC such that BC=hypotenuse, AB= Height and AC be the base.
Sol. Let AB be the tower and C and D be the points of observations. Then,
AB/AD=tan60®= √3 .
AD= AB/√3 =h/√3 .

AB/AC =tan 30®=1/√3 , AC=AB*√3 =h√3 .
CD= (AC-AD)=(h√3 -h/√3 ).
h√3 -h/√3=20.
h =10√3 =(10*1.73)=17.32
hence the height of tower is 17.32 meter.

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