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previous year questions of Height and Distance For NDA

previous year questions

previous year questions
Q 3220591411

The angle Of elevation of a stationary cloud from a point 25 m above a lake is 15° and the angle of depression of its image in the lake is 45° . The Height of the cloud above the lake level is
NDA Paper 1 2017
(A)

`25` m

(B)

`25 sqrt3 ` m

(C)

`50` m

(D)

`50 sqrt3` m

Solution:

`tan 15° = 2 – sqrt3 = x/(x + 50)`

`=> (2 – sqrt 3 –1 ) x = – 50(2 – sqrt3 )`

` => x = ( – 50(2 – sqrt3))/(1 - sqrt3)`

` => x + 25 = ( – 50(2 – sqrt3) + 25 (1 - sqrt3))/(1 - sqrt3)`

` = ( - 100 + 50 sqrt3 + 25 - 25 sqrt3)/(1 - sqrt3)`

` = ( - 75 + 25 sqrt3)/(1 - sqrt3) = ( 25(3 - sqrt3))/(sqrt3 - 1) = 25 sqrt3`
Correct Answer is `=>` (B) `25 sqrt3 ` m
Q 3220691511

The angles of elevation of he top of the tower from the top and foot of a pole ......? respectively 30° and 45° . If `h_T` is the height of the tower and `h_p` is the height of the pole, then which of the following are correct ?

1. `(2 h_P h_T)/(3 + sqrt3) = h_p^2`

2. `(h_T- h _P)/ ( sqrt3 + 1 ) = h_P/2`

3. `2 ( h_P + h_T)/h_P = 4 + sqrt3`

The select the correct answer using the cod given below.
NDA Paper 1 2017
(A)

1 and 3 only

(B)

2 and 3 only

(C)

1 and 2 only

(D)

1,2 and 3

Solution:

Let the distance between pole & tower is ‘b’.

Now , ` h_T/b = tan 45º = 1 => h_T = b`

` ( h_T - h_P)/b = tan 45º = 1/sqrt3 => ( h_T - h_P)/h_P = 1/sqrt3`

` => ( h_T - h_P)/( h_T - ( h_T - h_P) ) = 1/(sqrt3 - 1) => ( h_T - h_P)/h_P = ( sqrt3 + 1)/2`

`=>` Statement ‘`2`’ is correct,

` ( h_T - h_P + 2 h_p)/h_P = ( sqrt3 + 1 + 4)/2 => ( h_T +h_P)/h_P = (5 + sqrt3)/2`

`=>` Statement ‘`3`’ is incorrect.

`:. ` Option ‘`c`’ is right choice.
Correct Answer is `=>` (C) 1 and 2 only
Q 2711480320

The top of a hill when observed from the top and bottom of a building height h is at angles of elevation p and q respectively. What is the height of the hill ?
NDA Paper 1 2016
(A)

`(h cot q)/(cot q - cot p)`

(B)

`(h cot p) /( cot p - cot q)`

(C)

`(2h tan p)/( tan p - tan q)`

(D)

`(2 h tan q )/( tan q - tan p)`

Solution:

Let height of hill= H

Distance between hill & building `=x`

`:. triangle BCD` `tan P= (H-h)/x`

`x= (H-h) cot P`................(i)

`:. triangle ACO ` `tan q= H/x`

`x= H cot q`........................(ii)

from eqn (i) & (ii)

`(H-h) cot P= H cot q`

`h(cot p- cot q)= h cot p`

`H= (h cot p)/(cot p- cot q)`
Correct Answer is `=>` (B) `(h cot p) /( cot p - cot q)`
Q 2231101922

A vertical tower standing on a levelled field is mounted
with a vertical flag staff of length `3 m`. From a point on
the field, the angles of elevation of the bottom and tip of
the flag staff are `30^0` and `45^0`, respectively. Which one of
the following gives the best approximation to the height
of the tower?
NDA Paper 1 2015
(A)

`3.90 m`

(B)

`4.00 m`

(C)

`4.10 m`

(D)

`4.25 m`

Solution:

Let `AB` be tower of height `h m` and `BC` is the flag.

In `Delta BDA, tan 30^0 = h/x => x = sqrt(3) h` .......(i)

Now,in `DeltaCDA, tan 45^0 = (h+3)/x`

` => x = h + 3 => h = x - 3 = sqrt(3) h - 3` [From Eq. (i)]

`=> h ( 1- sqrt(3) ) -3`

`:. h = 3/(sqrt(3)-1) = (3 (sqrt(3) +1))/2 = 1.5 xx 2.732`

`= 4.098 m = 4.10 m` (approx.)
Correct Answer is `=>` (C) `4.10 m`
Q 1648167003

The angle of elevation of the top of a tower from a
point `20 m` away from its base is `45^0`. What is the
height of the tower?
NDA Paper 1 2015
(A)

`10 m`

(B)

`20 m`

(C)

`30 m`

(D)

`40 m`

Solution:

In `Delta ABC, tan45^0 = h/(20) => h =20m`
Correct Answer is `=>` (B) `20 m`
Q 1628267101

The angles of elevation of the top of a tower
standing on a horizontal plane from two points on
a line passing through the foot of the tower at
distances `49 m` and `36 m` are `43^0` and `47^0`
respectively. What is the height of the tower?
NDA Paper 1 2015
(A)

`40 m`

(B)

`42 m`

(C)

`45 m`

(D)

`47 m`

Solution:

In ` Delta ACD`,

` tan47^0 = h/(36) ` .........(1)

and in ` Delta ABD , tan43^0 = h/(49)` .............(2)

` => 1/ (cot 43^0) = h/(49)`

` => (49)/ (cot 43^0) = h`

` => h = (49)/(cot ( 90 - 47 ^0)`

` => h = (49)/(h/(36)`

` => h^2 = 49 xx 36 `

`:. h = 7 xx 6 = 42 m`
Correct Answer is `=>` (B) `42 m`
Q 1780801717

A lamp post stands on a horizontal plane. From a point
situated at a distance `150 m` from its foot, the angle of
elevation of the top is `30^0`. What is the height of the
lamp post?
NDA Paper 1 2014
(A)

`50 m`

(B)

`50 sqrt(3) m`

(C)

`(50)/ sqrt(3) m`

(D)

`100 m`

Solution:

Let `AB` be the lamp post of height hand `C` is

a point situated at a distance of `150 m` from its foot `B`.

In `Delta ABC`, we have

`tan 30^0 = h/(150)`

`=> 1/sqrt(3) = h/(150)`

` :. h = (150) / sqrt(3) m`

` = (150 xx sqrt(3))/3 = 50sqrt(3) m`
Correct Answer is `=>` (B) `50 sqrt(3) m`
Q 2319856719

If the angles of elevation of the top of a tower
from two places situated at distances `21 m` and `x`
`m` from the base of the tower are `45°` and `60°`
respectively, then what is the value of `x`?
NDA Paper 1 2013
(A)

`7 sqrt 3`

(B)

`7-sqrt 3` m

(C)

`7+sqrt 3 `m

(D)

`14` m

Solution:

Let h be the height of the tower.

In `Delta ACD, tan 45^o=h/21`

`=> 1=h/21`

`:. h=21 m`............(1)

Again, in `Delta ABD`

`tan 60^o=h/x`

`=> sqrt 3=h/x`

`:. x=21/sqrt 3=7 sqrt 3 m` [fromEq.(i)]
Correct Answer is `=>` (A) `7 sqrt 3`
Q 2329167011

The angle of elevation of the top of a tower of
height H from the foot of another tower in the
same plane is `60°` and the angle of elevation of the
top of the second tower from the foot of the first
tower is `30°`. If h is the height of the other tower,
then which one of the following is correct?
NDA Paper 1 2013
(A)

`H = 2 h`

(B)

`H = sqrt 3 h`

(C)

`H = 3 h`

(D)

None of these

Solution:

In `Delta DBC`,

`tan 60^o=H/(BC)=> sqrt 3=H/(BC)=> BC=H/(sqrt 3)`............(i)

In `Delta ACB`

`tan 30^o=h/(BC)=> 1/(sqrt 3)=> BC=h sqrt 3`................(ii)

From Eqs. (i) and (ii),

`H/(sqrt 3)= h sqrt 3`

`H= 3h`
Correct Answer is `=>` (C) `H = 3 h`
Q 2359467314

A man walks `10 m` towards a lamp post and
notices that the angle of elevation of the top of the
post increases from `30°` to `45°`. The height of the
lamp post is
NDA Paper 1 2013
(A)

`10` m

(B)

`(5 sqrt 3 +5)m`

(C)

`(5 sqrt 3-5) m`

(D)

`(10 sqrt 33+10 ) m`

Solution:

Let `BL = x m` and `PL = h m`

Now, in `Delta PBL`,

`tan 30^o=h/(10+x)=1/sqrt 3`

`=>1/sqrt 3 =h/(10+x)=> sqrt 3 h=10 +x`

`=> (sqrt 3-1)h=10`

`:. h=(10)/(sqrt 3-1) *(sqrt 3+1)/(sqrt 3+1)`

`= (10 (sqrt 3+1))/(3-1)=(10 (sqrt 3+1))/2`

`=5(sqrt 3+1)=(5 sqrt 3+5)m`
Correct Answer is `=>` (B) `(5 sqrt 3 +5)m`
Q 2319467319

The shadow of a tower standing on a level plane is
found to be `50 m` longer when the Sun's elevation
is `30°` than when it is `60°`. The height of the
tower is
NDA Paper 1 2013
(A)

`25 m`

(B)

`25 sqrt 3` m

(C)

`50 m`

(D)

None of these

Solution:

Let shadow of a tower made by a Sun at an `angle 60°` is
`x m`. Then, by given condition, The shadow of a tower made by a
Sun at an `angle 30°` is `50 m` longer than at an angle `60°` .i.e., `(50+x)m`.

Now, in `Delta ACB, tan 60^o=h/x=sqrt 3=> x=h/sqrt 3 m`...............(i)

and in `Delta ADB`

`tan 30^o=h/(x+50)=1/sqrt 3=> sqrt 3h=x+50`

`=> sqrt 3 h=h/sqrt 3+50` [from Eq. (i)]

`=> (sqrt 3-1/sqrt 3)h=50=> (3-1)/(sqrt 3) h=50=> 2h=50 sqrt 3`

`h=25 sqrt 3`
Correct Answer is `=>` (B) `25 sqrt 3` m
Q 2379567416

The top of a hill observed fron1 the. top and
bottom of a building of height h is at angles of
elevation `alpha` and `beta`, respectively. The height of the
hill is
NDA Paper 1 2012
(A)

`(h cot beta)/(cot beta-cot alpha)`

(B)

`(h cot alpha)/(cot alpha-cot beta)`

(C)

`(h cot beta)/(cot beta+cot alpha)`

(D)

None of these

Solution:

Let `AD = BE = y` and `AB = DE = h`

Now, in `triangle CAD` `tan alpha=x/y=> y=x cot alpha`.............(i)

Again in `triangle CBE`

`tan beta=(x+h)/y`

`=> y=(x+h) cot beta`.............(ii)

From Eqs. (i) and (ii), `xcot alpha = (x + h)cot beta`

`=> xcot alpha = xcot beta + h cot beta`

`=>x (cot alpha-cot beta)=h cot beta=> x=(h cot beta)/(cot alpha-cot beta)`..............(iii)

`:.` :. Required height `(CE) = CD + DE = x + h`

`=(h cot beta)/(cot alpha-cot beta)+h=(h cot alpha)/(cot alpha-cot beta)`
Correct Answer is `=>` (B) `(h cot alpha)/(cot alpha-cot beta)`
Q 2319667519

The angle of elevation of a tower at a level ground
is `30°`. The angle of elevation becomes `theta` when
moved `10 m` towards the tower. If the height of
tower is `5 sqrt 3 m`, then what is the value of `theta`?
NDA Paper 1 2012
(A)

`45^o`

(B)

`60^o`

(C)

`75^o`

(D)

None of these

Solution:

In `Delta DBC`

`than theta=(5 sqrt 3)/x`...............(i)

Again , in `Delta DAC`

`tan 30^o =(5 sqrt 3)/(10+x)=1/(sqrt 3) => 10+x=15`

` :. x=5 m`

From Eq. (i),

`tan theta=(5 sqrt 3)/5=sqrt 3=tan 60^o=> theta=60^o`
Correct Answer is `=>` (B) `60^o`
Q 2369767615

The angle of elevation of the tip of a flag staff
from a point `10 m` due South of its base is `60^o`
What is the height of the flagstaff correct to the
nearest metre?
NDA Paper 1 2012
(A)

`15 m`

(B)

`16 m`

(C)

`17 m`

(D)

`18 m`

Solution:

In `triangle BAC`,

`tan 60^o=h/10`

`:. h=10 tan 60^o=10 * sqrt 3`

`=10 * (1.732)=17.32 m=17 \ \ m` (approx)
Correct Answer is `=>` (C) `17 m`
Q 2319867710

A tower of height `15 m` stands vertically on the
ground. From a point on the ground the angle of
elevation of the top of the tower is found to be
`30°`. What is the distance of the point from the
foot of the tower'?
NDA Paper 1 2011
(A)

`15 sqrt 3 m`

(B)

`10 sqrt 3 m`

(C)

`5 sqrt 3 m`

(D)

`30 m`

Solution:

Let `AB=x m`

Then, in `Delta BAC`

`tan 30^o=(BC)/(AB)=15/x`

`=> 1/sqrt 3=15/x`

`:. x=15 sqrt 3 m`
Correct Answer is `=>` (A) `15 sqrt 3 m`
Q 2349867713

At a point `15 m` away from the base of. a `15 m` high
house, the angle of elevation of the top is
NDA Paper 1 2011
(A)

`90^o`

(B)

`60^o`

(C)

`45^o`

(D)

`30^o`

Solution:

Let the angle of elevation = `theta`

In `triangle BAC`

`tan theta=(BC)/(AB)=15/15=1`

`=> tan theta=tan 45^o`

`=> theta=45^o`
Correct Answer is `=>` (C) `45^o`
Q 2309867718

A vertical tower stands on a horizontal plane and
is surmounted by a vertical flag-staff:of height `h`.
At a point `P` on the plane, the angle of,elevation of
the bottom of the flag-staff is `beta` and that of the top
is `alpha`. What is the height of the tower'?
NDA Paper 1 2011
(A)

`(h tan beta)/(tan alpha-tan beta)`

(B)

`(h tan beta)/(tan alpha+tan beta)`

(C)

`(h cos beta)/(cos alpha-cos beta)`

(D)

`h/(cos(alpha-beta))`

Solution:

Let the height of the tower `= y = BC`

Now In `Delta CPB`

`tan beta=y/x`..............(i)

and in `Delta APB`

`tan alpha=(AB)/(BP)=(AC+BC)/(BP)`

`=> tan alpha=(h+y)/x`

`=> y+h= x tan alpha`

`=> y+h=y/(tan beta) * tan alpha` from Eq. (i)]

`=> y(1-(tan alpha)/(tan beta))=-h`

`=> y(tan alpha-tan beta)=h tan beta`

`:.` Heigh of the tower `y=(h tan beta)/((tan alpha-tan beta))`
Correct Answer is `=>` (A) `(h tan beta)/(tan alpha-tan beta)`
Q 2349167913

An aeroplane flying at a height of ` 300 m` above the
ground passes vertically above another plane at an
instant when the angles of elevation of two planes
from the same point on the ground ,are `60°` and
`45°`, respectively. What is the height of the lower
plane from the ground'?
NDA Paper 1 2011
(A)

`50 m`

(B)

`100/(sqrt 3) m`

(C)

`100 sqrt 3 m`

(D)

`150 (sqrt 3+1) m`

Solution:

Let the height of the lower plane from the ground `= x`

and `PA=y`

Now, In `Delta APB`

`tan 45^o=x/y=(AB)/(AP)=1=> x=y`..............(i)

and in `Delta APC`

`tan 60^o=(AC)/(AP)=300/y=sqrt 3`

`=> y=300/sqrt 3`

`:. x=300/sqrt 3 * sqrt 3/sqrt 3=(300 sqrt 3)/3` [fromEq.(i)]

`=100 sqrt 3`
Correct Answer is `=>` (C) `100 sqrt 3 m`
Q 2369067815

A man standing on the bank of a river observes
that the angle of elevation of the top of a tree just
on the opposite bank is `60°`. The angle of elevation
is `30°` from a point at a distance `y m` from the
bank. What is the height of the tree?
NDA Paper 1 2011
(A)

`y m`

(B)

`2y m`

(C)

`(sqrt 3 y)/2 m`

(D)

`y/2 m`

Solution:

Let `BC = x m`

In `Delta ACD`

`tan 30^o=(CD)/(AC)=> 1/sqrt 3=h/(x+y)`

`=> x+y=h sqrt 3`............(i)

and in `Delta CBD`

`tan 60^o=(CD)/(BC)=> sqrt 3=h/x`

`=> x=h/sqrt 3`.....................(ii)

From Eqs. (i) and (ii),

`h/sqrt 3+y =h sqrt 3=>
y=h (sqrt 3-1/sqrt 3)=(2h)/sqrt 3`

`:. h=(sqrt 3y)/2 m`
Correct Answer is `=>` (C) `(sqrt 3 y)/2 m`
Q 2349380213

The angle of elevation of the top of a flag post
from a point `5 m` away from its base is `75°`. What
is the approximate height of the flag post?
NDA Paper 1 2010
(A)

`15 m`

(B)

`17 m`

(C)

`19 m`

(D)

`21 m`

Solution:

Let `h` be the height of the flag post

In `Delta ACB`

`tan 75^o=(AB)/(BC) =h/5`

`=> tan(45^o+30^o)=(tan 45^o +tan 30^o)/(1-tan 45^o tan 30^o)=h/5`

`=> (1+sqrt 3)/(sqrt 3-1)=h/5`

`:. h=((sqrt 3+1)^2)/((sqrt 3)^2-(1)^2) xx 5`

`=5((3+1+2 sqrt 3)/(3-1))`

`=5(2+sqrt 3)`

`=5 xx 3.732`

`=18.660`

`=19 m`
Correct Answer is `=>` (C) `19 m`
Q 2349580413

A man observes the elevation of a balloon to be
`30°`. He, then walks `1 km` towards the balloon and
finds that the elevation is `60°`. What is the height
of the balloon?
NDA Paper 1 2009
(A)

`1//2 km`

(B)

`sqrt 3 //2 km`

(C)

`1//3 km`

(D)

`1 km`

Solution:

In `Delta CBD`

`tan 60^o =(CD)/(BC)`

`=> sqrt 3=h/y`

`=> h=sqrt 3 y`.............(i)

and in `Delta CAD`

`tan 30^o =(CD)/(AC)`

`=> 1/sqrt 3=h/(1+y)`

`=> 1+y=h sqrt 3`

`=> 1+y=3y` [from Eq. (i)]

`=> y=1/2`

`:.` Required height `(h)= sqrt3/2 km` [from Eq. (i)]
Correct Answer is `=>` (B) `sqrt 3 //2 km`
Q 2389580417

The angle of elevation from a point on the bank of
a river of the top of a temple on the other bank is
`45°`. Retreating `50 m`, the observer finqs the new
angle of elevation as 30°. What is the width of the
river?
NDA Paper 1 2009
(A)

`50 m`

(B)

`50 sqrt 3 m`

(C)

`50//(sqrt 3-1) m`

(D)

`100 m`

Solution:

In `Delta ACB`

`tan 45^o=(AB)/(BC)`

`=> 1=h/x`

`=> h=x`

Now, in `Delta ADB`

`tan 30^o =(AB)/(BD)`

`=> 1/sqrt 3=h/(x+50)`

`=> x+50 =h sqrt 3`

`=> h+50=h sqrt 3` [from Eq. (i)]

`=> h=50/(sqrt 3-1)`

`:.` Width of the river `(x) = h =50/(sqrt 3-1) m`
Correct Answer is `=>` (C) `50//(sqrt 3-1) m`
Q 2309680518

The foot of a tower of height h m is in a direct line
between two observers `A` and `B`. If the angles of
elevation of the top of the tower as seen from `A`
and `B` are a and p respectively and if `AB = d m,`
then what is `h // d` equaf to?
NDA Paper 1 2008
(A)

`(tan(alpha+beta))/((cot alpha cot beta-1))`

(B)

`(cot(alpha+beta))/((cot alpha cot beta-1))`

(C)

`(tan(alpha+beta))/((cot alpha cot beta+1))`

(D)

`(cot(alpha+beta))/((cot alpha cot beta+1))`

Solution:

Let `AD=x`

`:. DB=d-x`

In `Delta CAD, tan alpha=h/x=> x=h cot alpha`.................(i)

In `Delta CBD, tan beta=h/(d-x)=> d-x=h cot beta`.............(ii)

From Eqs. (i) and (ii),

`d=h(cot alpha+cot beta)`................(iii)

We know that,

`cot(alpha+beta)=(cot alpha cot beta-1)/(cot beta+cot alpha)`

`=> cot beta+cot alpha=(cot alpha cot beta-1)/(cot(alpha+beta))`

`:. d=h[(cot alpha cot beta-1)/(cot (alpha+beta))]` [from Eq. (iii)]

`=> h/d=(cot(alpha+beta))/(cot alpha cot beta-1)`
Correct Answer is `=>` (B) `(cot(alpha+beta))/((cot alpha cot beta-1))`
Q 2319780619

PT, a tower of height `2^ x m`, `P` being the foot, ` T`
being the top of the tower. `A, B` are points on the
same line with `P` If `AP = 2 ^(x+ 1) m, BP = 192 m` and
if the angle of elevation of the tower as seen from
`B` is double the angle of the elevation of the tower
as seen from `A`, then what is the value of `x` ?
NDA Paper 1 2008
(A)

`6`

(B)

`7`

(C)

`8`

(D)

`9`

Solution:

Here, given that `PT=2^x m, AP=2^(x+1) m` and `BP=192 m`

In `Delta PAT, ` `tan theta =(PT)/(AP)`

`=> tan theta=2^x/(2^(x+1))=1/2`................(1)

Now, in `Delta PBT`

`tan 2 theta=(PT)/(PB)=2^x/192`

`=> (2 tan theta)/(1-tan^2 theta)=2^x/192`

`=> (2 1/2)/(1-1/4)=2^x/192`

`=> 4/3 xx 192 =2^x` (from 1)

`=>2^x=256`

`=> 2^x=2^8`

`:. x=8`
Correct Answer is `=>` (C) `8`
Q 2319880719

What should be the height of a flag where a `20` ft
long ladder reaches `20` ft below the flag (The
angle of elevation of the top of the flag at foot
of the ladder is `60°` )?
NDA Paper 1 2007
(A)

`20` ft

(B)

`30` ft

(C)

`40` ft

(D)

`20 sqrt 2` ft

Solution:

Let `BD` is a flag and `80 = (h + 20)` ft.

In `Delta BAD`

`tan 60^o=(BD)/(AB)`

`=> sqrt 3=(h+20)/(AB)`

`=> AB=((h+20))/sqrt 3`....................(i)

Now, in `Delta ABC`

`AC^2=AB^2+BC^2` (by Pythagoras theorem)

`=> 400 =((h+20)^2+3h^2)/3`

`=> 1200=h^2+40h+400+3h^2`

`=> 4h^2+40 h-800=0`

`=> h^2+10h-200=0`

`=>h^2+20 h-10h-200=0`

`=> h(h+20)-10(h+20)=0`

`=>h=10`

`:.` height of flag `(BD)=BC+CD=h+20`

`=10+20=30 `ft
Correct Answer is `=>` (B) `30` ft
 
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