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.
• Distance between the boats A and B is x
• Distance between the boats B and C is y.

To find :

The height of the light-house.

Solution :

Let the height of the light house be h.

From the Attachment ( Figure )

• PR = h
• < R= 90°
• AB = x and BC = y

In ∆PBC.

∠RCP= ∠CBP +∠CPB [ exterior angle theorem ]

∠CPB=3a-2a = a

now ,In ∆PAB

∠CBP=∠BAP+∠BPA [exterior angle theorem]

∠BPA=2a-a= a

Now,In ∆PAB

∠BPA=∠BAP

⇒PB= AB= x .....(1)[ side opposite to equal angles of a triangle are equal ]

Now , In ∆ PRC

$\sf\sin3a=\dfrac{h}{CP}$

$\sf\:h=\dfrac{\sin3a}{CP}...(2)$

Similarly In ∆APR

$\sf\sin\:a=\dfrac{h}{AP}$

$\sf\:h=\dfrac{\sin\:a}{AP}...(3)$

In ∆CPA

BM is a angular bisector

$\sf\dfrac{CP}{y}=\dfrac{AP}{x}....(4)$

From equation (2)&(3)

$\sf\dfrac{\sin\:a}{AP}=\dfrac{\sin3a}{CP}$

$\sf\dfrac{\sin3a}{\sin\:a}=\dfrac{AP}{CP}$

Form equation (4)

$\sf\dfrac{\sin3a}{\sin\:a}=\dfrac{x}{y}$

We know that sin3a=3 sina - 4 sin³a

$\sf\implies\dfrac{x}{y}=3-\sin^2a$

$\sf4\sin^2a=3-\dfrac{x}{y}..(4)$

We know that cos2a=1-2sin²a

Multiply both sides by 2

2cos2a=2-4sin²a

From equation (4)

$\sf2\cos2a=2-(3-\dfrac{x}{y})$

$\sf\implies2\cos2a=\dfrac{x}{y}-1$

$\sf\implies\cos2a=(\dfrac{x-y}{2y})$

Now

$\sf\sin2a=\sqrt{1-cos^2\:2a}$

$\sf\sin2a=\sqrt{1-(\frac{x-y}{2y})^2$

$\sf\sin2a=\sqrt{\dfrac{4y^2-x^2-y^2+2xy}{y^2}}$

$\sf\sin2a=\dfrac{\sqrt{3y^2-x^2+2xy}}{2y}$

$\sf\sin2a=\dfrac{\sqrt{(x+y)(3y-x)}}{2y}$

In ∆BPR

$\sf\sin2a=\dfrac{h}{BP}$

$\sf\:h=BP\sin2a$ [ from equation (1)]

$\sf\:h=x\sin2a$

$\sf\:h=x\times\dfrac{\sqrt{(x+y)(3y-x)}}{2y}$

Therefore, Height of the light house is

$\sf\:=x\times\dfrac{\sqrt{(x+y)(3y-x)}}{2y}$