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.


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Answer 1

✰ 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.

✪ 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}$

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

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

➙ $\Large\tt sin^2a=\frac{3y-x}{4y}$

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

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

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

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

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

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

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

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

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

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

Answer 2

ANSWER:-

here, <Q = 90, PQ = h, AB = X So, IN A BPO,

tan2a = PQ/QB

QB = h/tan2a- -(1)

now, IN APQC,

tan3a = PQ/QC

QC = h/tan3a-

similarly, IN AAPQ tana = PQ/(QA)

(2)

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

(QC + BC + x) = h/tana ( AB = x

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

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

Form- (1),& (3).

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

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

tanaſh + xtan2a] = stanza

tana + xtana.tan2a = tana

h(tan2a - tana) = xtana.tan2a

$thanks$