Question

The angles of elevation of the top of a lighthouse from 3 boats A , B and C in a straight line of same side of the light house are a , 2a , 3a respectively . If the distance between the boats A and B and the boats B and C are x and y respectively find the height of the light house ?

Answer 1

Given :  The angles of elevation of the top of a lighthouse from 3 boats A , B and C in a straight line of same side of the light house are a , 2a , 3a respectively  .   distance between the boats A and B and the boats B and C are x and y respectively

To find :  Height of light house in terms of x & y

Solution:

let say top of  lighthouse =  X

base of  lighthouse = O

height of lighthouse = h

∠OCX = 3a    ∠OBX = 2a  ∠OAX =  a

∠OCX =  ∠OBX +  ∠CXB   => ∠CXB  = a

similarly  ∠BXA  = a

in Δ XBC

∠BXA  = ∠BAX = a

=> BX  = AB

=> BX  =  x  

Sin2a  =  h /BX

=> Sin2a = h/x

=> h = xSin2a

now in Δ CXA

BX is angle bisector

=> CX/y  =  AX/x

CX = h/Sin3a  & AX   = h/Sina

=>  h/ysin3a  =  h/xsina

=>   x/y  =  sin3a /sina

=> x/y  =   (3 sina - 4 sin³a)/sina

=>  x/y  =  3 - 4Sin²a

Cos2a  = 1 - 2sin²a    => -4sin²a  =  2Cos2a   - 2

=> x/y  =   3 + 2Cos2a   - 2

=>2Cos2a  = x/y   - 1

=> Cos2a  =  (x - y)/2y

Sin²2a =  1 - Cos²2a

=> Sin²2a =   1  -  ((x - y)/2y)²

=> Sin²2a =   ( (x + y)(3y - x)  ) /(2y)²

=> Sin 2a =   √ (x + y)(3y - x)    / 2y

h = xSin2a

=> h = x   √((x + y)(3y - x) )   / 2y  

Height of light House =   x  √((x + y)(3y - x))  / 2y  

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Answer 2

ANSWER: -

$h = x \sqrt{((x + y)(3y - x)) \div 2y}$

SOLUTION: -

GIVEN :- The angles of elevation of the top of the light house from 3boats A,B,C in a straight line of same side of the light house are a,2a,3a respectively. And Distance between the boats A&B and the boats B&C are x&y respectively.

REQUIRED TO FIND :- Height of lighthouse .

CALCULATION: -

From the figure,

In Triangle XBA,

/_BXA = /_BAX =a

=>BX=AB

AND,

=> sin2a = h /BX

sin2a = h/x

=> h = xsin2a. -------(1)

In triangle CXA,

● BX is an angle bisector.

=>CX/ y = AX /x

=>CX = h/sin3a = h/xsina

=>x/y = sin3a / sina

=>x/y = (3sina - 4sin^3 a) / sina

=>x/y= 3 -4 sin^2 a

And,

☆ Cos 2a = 1 - sin^2 a

=> -4sin^2 a = 2cos2a-2

=>x/y = 3+2cos2a -2

=>2cos2a = x/y -1

=> cos 2a = (x-y)/2y

Also,

☆ sin^2 2a = 1 - cos ^2 2a

=> sin^2 2a = 1-cos ^2 2a

=> sin^2 2a = 1 - [ (x-y) /2y ] ^2

=> [ (x+y)(3y-x) /2y ]^2

=> sin2a =

$\sqrt{(x + y)(3y - x) \div 2y}$

And from (1) : h = xsin2a

So,

$h = \sqrt{(x + y)(3y - x) \div 2y}$

HOPE IT HELPS :)