Math, asked by offfff53, 10 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

14

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here, <Q = 90, PQ = h , AB = x

So, IN ∆ BPQ,

tan2a = PQ/QB

QB = h/tan2a----------( 1 )

now, IN ∆PQC,

tan3a = PQ/QC

QC = h/tan3a-----------( 2 )

similarly, IN ∆APQ

tana = PQ/(QA)

tana = h/(QC + BC + AB) [ as QA = QC + BC + AB]

(QC + BC + x) = h/tana [ AB = x]----------( 3 )

We may write [BC = QB - QC] we get,

[ QC + QB - QC + X] = h/tana

From-------( 1 ) , & -------( 3 ).

[h/tan 2a + x] = h/tana

[(h + xtan2a)/tan2a] = h/tana

tana[h + xtan2a] = htan2a

htana + xtana.tan2a = htan2a

h(tan2a - tana) = xtana.tan2a

h=xtana.tan2a/tan2a - tana

Answered by Anonymous

20

✰ Qúēsᴛíõɴ :-

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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✪ Sōʟúᴛîôñ :-

$\\$

Let PQ be the height of light house = h m

$\\$

A = 1st point of observation
B = 2nd point of observation
C = 3rd point of observation

$\\$

Given, AB = x and BC = y $\\$

Exterior angle = Sum of opposite interior angles $\\$

∠PBQ = ∠BQA + ∠BAQ and

∠PCQ = ∠CBQ + ∠CQB $\\$

⛬ AB = x = QB

By applying the sine rule,

$\\$

∆BQC we get

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

$\\$

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

$\\$

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

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

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➙ $\Large\tt 4sin^2a=3-$ $\Large\tt\frac{x}{y}=\frac{3y-x}{y}$

$\\$

➙ $\Large\tt sin^2a=\frac{3y-x}{4y}$

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

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

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$\Large\tt from\:\:\triangle{PBQ,}$

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

$\\$

$\Large\tt ⇒4sin^2a\:cos^2a=\frac{h^2}{x^2}$

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

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$\Large\tt⇒h^2=\frac{x^2}{4y^2}(3y-x)(x+y)$

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

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$\Large\tt\therefore$ $\Large\tt Height\:of\:lighthouse$

$\\$

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

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