Answered

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation f his eyes to the top of the building increases from 30' to 60' as he walks towards the building. Find the distance he walked towards the building.

Answer :

LET ,

HEIGHT OF THE TOWER(CE)=30 M

HEIGHT OF THE BOY(AB)=1.5M

⇒CD=AB=1.5 M

⇒DE=CE-CD=30-1.5

                       =28.5

ANGLE OF ELEVATION

ANGLE EAD= 30 DEG  ,ANGLEEFD=60 DEG

FROM,

TRIANGLE EAD,

TAN 30 =DE/AD

1/√3=28.5/X+Y

X+Y=28.5√3.........................EQ(1)

FROM,TRIANGLE EFD,

TAN 60=ED/FD

√3=28.5/Y

Y=28.5/√3

SUB Y =2S.5/√3 IN EQ(1)

X+28.5/√3=28.5√3

X=28.5√3-28.5/√3

X=28.5(√3-1/√3)

X=28.5×2/√3×√3/√3

X=(57×√3)√3

X=19×√3

X=19(1.732)

X=33.52

THEREFORE DISTANCE TRAVELLED BY BOY IS 33.908
HEIGHT OF THE TOWER(CE)=30 MHEIGHT OF THE BOY(AB)=1.5M⇒CD=AB=1.5 M⇒DE=CE-CD=30-1.5                       =28.5ANGLE OF ELEVATIONANGLE EAD= 30 DEG  ,ANGLEEFD=60 DEGFROM,TRIANGLE EAD,TAN 30 =DE/AD1/√3=28.5/X+YX+Y=28.5√3.........................EQ(1)FROM,TRIANGLE EFD,TAN 60=ED/FD√3=28.5/YY=28.5/√3SUB Y =2S.5/√3 IN EQ(1)X+28.5/√3=28.5√3X=28.5√3-28.5/√3X=28.5(√3-1/√3)X=28.5×2/√3×√3/√3X=(57×√3)√3X=19×√3X=19(1.732)X=33.52THEREFORE DISTANCE TRAVELLED BY BOY IS 33.908

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