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 lighthouse.​

Answer 1

$\huge{\orange{\underline{\red{\mathbf{QUESTION}}}}}$

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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 lighthouse.

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$\huge{\orange{\underline{\red{\mathbf{SOLUTION}}}}}$

Given, AB = x and BC = y

Exterior angle = Sum of opposite interior angles

∠PBQ = ∠BQA + ∠BAQ and

∠PCQ = ∠CBQ + ∠CQB

$\therefore$AB = x = QB

By applying the sine rule,

∆BQC we get

$\frac{BQ}{sin∠QCB}=$$\frac{BC}{sin∠CQB}$

$⇒\frac{x}{sin(180°-3a)}=$$\frac{y}{sin\:a}$

$⇒\frac{x}{y}=$$\frac{sin3a}{sin\:a}=$$\frac{3sin\:a-4sin^3a}{sin\:a}$

$\:\:\:\:\:\:\:=3-4sin^2a$

$⇒4sin^2a=3-$$\frac{x}{y}=\frac{3y-x}{y}$

$⇒sin^2a=\frac{3y-x}{4y}$

$\therefore$$cos^2a=1-sin^2a=1-\frac{3y-x}{4y}$

$\:\:\:\:\:\:\:\:\:=\frac{4y-3y+x}{4y}=\frac{y+x}{4y}$

$from\triangle{PBQ,}$

$sina=\frac{h}{x}⇒2sinacosa=\frac{h}{x}$

$⇒4sin^2a\:cos^2a=\frac{h^2}{x^2}$

$⇒4\:.\frac{3y-x}{4y}\:.\frac{x+y}{4y}=\frac{h^2}{x^2}$

$⇒h^2=\frac{x^2}{4y^2}(3y-x)(x+y)$

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

$\therefore$$Height\:of\:lighthouse$

$\:\:\:\:\:=\frac{x}{2y}\sqrt{(3y-x)(x+y)}\:m$