Math, asked by patkarirekha, 11 months ago

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.​

Answers

Answered by Anonymous
55

\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

\thereforeAB = 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}

\thereforecos^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)

\thereforeh=\frac{x}{2y}\sqrt{(3y-x)(x+y)}

\thereforeHeight\:of\:lighthouse

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

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