AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$A$ tower standing at point $A$ leans towards west making an angle $α$ with the vertical.

The angular elevation of $B$ , the top most point of the tower is $β$ as observed from a point $C$ due east of $A$ at a distance $d$ from $A$ .

lf the angular elevation of $B$ from a point due east of $C$ at a distance $2 d$ from $C$ is $γ$ , then $2 tanα$ can be written as

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

cotβ - 2 cotγ

cotγ - cotβ

3 cotβ - cotγ

cotβ - 3 cotγ

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

It is given a tower standing at point

A

leans towards west making an angle

α

with the vertical.
The angular elevation of

B

, is

β

as observed from the point

C

which is at a distance

d

from

A

in the direction east.
According to the given data, the figure has been drawn below.
Question Image

Let

AB,

be the tower leaning towards west making an angle

α

with vertical, at

C

angle of elevation of

B

β

and at

D

γ

.
Let BH=h.

CA = AD = d

△ ABH

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

tanα = \frac{AH}{BH}

\Rightarrow tanα = \frac{AH}{h}

\Rightarrow AH = htanα . . . . . . . . . (1)

△ BCH,

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

tanβ = \frac{BH}{CH}

\Rightarrow tanβ = \frac{h}{CH}

\Rightarrow CH = hcotβ

∵ 1 tan θ = cot θ

Now from the figure,

CA = CH + AH

\Rightarrow CH = CA - AH

\Rightarrow CH = d - AH

Substituting

CH = hcotβ

∴

d - AH = hcotβ,

Put the value of AH from (

1)

d - htanα = hcotβ

\Rightarrow d = h (cotβ + tanα) . . . . . . . (2)

△ BDH,

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

tanγ = \frac{BH}{HD}

\Rightarrow \frac{h}{HA + AD} = tanγ

\Rightarrow AH + d = \frac{h}{tanγ}

\Rightarrow AH + d = hcotγ

∵ 1 tan θ = cot θ

\Rightarrow htanα + d = hcotγ

(from the equation 1)

\Rightarrow d = h (cotγ - tanα) . . . . . . . . . . . . . (3

)
From

(2)

and (

3)

we get,

h (cotβ + tanα) = h (cotγ - tanα)

\Rightarrow cotβ + tanα = cotγ - tanα

\Rightarrow 2 tanα = cotγ - cotβ

So,

2 tanα

can be written as

cotγ - cotβ

.
So, the correct option is 2.

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!