Question

Question #7c5e8

Answer 1

In the figure

`AB` is the mast of the ship of height `h`.

The `3/4th` part of `AB` i.e.`AD=(3h)/4` subtends an angle `a` at C , a point 40m from the mast i.e `BC=40`m.

Let us consider that lower `h/4=BD` subtends angle `b` at C.

So `(AB)/(BC)=tan(a+b)`

`=>h/40=tan(a+b).......[1]`

Again `(BD)/(BC)=tanb`

`=>(h/4)/40=tanb`

`=>h/160=tanb......[2]`

Dividing [1] by [2] we get

`4=tan(a+b)/tanb`

`=>4tanb=(tana+tanb)/(1-tanatanb)`

given `tana =3/4` and let `tanb=x`

we get

`4x=(3/4+x)/(1-3/4x)`

`=>4x=(3+4x)/(4-3x)`

`=>16x-12x^2=3+4x`

`=>12x^2-12x+3=0`

`=>4x^2-4x+1=0`

`=>(2x-1)^2=0`

`=>x=1/2`

`=>tanb=1/2`

Using [2] we have

`h=160xxtanb=160xx1/2=80` m