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$PQ$ is a vertical tower. $A, B, C$ are three points in a horizontal line through $Q$ , the foot of the tower. If the angles of elevation of the top of the tower from $A, B, C$ are $3 θ$ , $2 θ, θ$ respectively, then $BCcot 3 θ - CAcot 2 θ + ABcotθ$ is equal to

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3

1

2

0

answer is D.

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Detailed Solution

It is given that

PQ

is a vertical tower and

A, B, C

are three points in a horizontal line through the foot of the tower.
The angles of elevation of the top of the tower from

A, B, C

are

3 θ

2 θ, θ

respectively.
According to the given data, the figure has been drawn below.
Question Image

Now,
In

Δ A P Q,

cot θ = b a s e p e r p e n d i c u l a r

cot 3 θ = Q A P Q

……(1)
In

Δ B P Q,

cot θ = b a s e p e r p e n d i c u l a r

cot 2 θ = Q B P Q

……(2)
In

Δ C P Q,

cot θ = b a s e p e r p e n d i c u l a r

cot θ = Q C P Q

……(3)
From eq. (1), (2) and (3) substitute the values in the given expression, we get

BCcot 3 θ - CAcot 2 θ + ABcotθ

\Rightarrow \frac{BC \times QA}{PQ} - \frac{CA \times QB}{PQ} + \frac{AB \times QC}{PQ}

From the given figure,

BC = QC - QB;

CA = QC - QA;

AB = QB - QA

.
Now replacing BC , CA and AB as the expressions simplified above, we get the following expressions.

\Rightarrow \frac{QA (QC - QB)}{PQ} - \frac{(QC - QA) QB}{PQ} + \frac{(QB - QA) QC}{PQ}

\Rightarrow \frac{1}{PQ} [QCQA - QB QA - QC QB + QA QB + QB QC - QA QC]

\Rightarrow \frac{1}{PQ} \times 0

\Rightarrow 0

So, we get

BCcot 3 θ - CAcot 2 θ + ABcotθ = 0

.
So, the correct option is 4.

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