Mathematics Tricks & Tips Of Height and Distance For NDA
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Tricks & Tips Of Height and Distance For NDA

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Calculation of height and distance : Elevation angel is given

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 2379367216

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
Paper 1 2013
(A)

`10` m

(B)

`(5 sqrt 3 +5)m`

(C)

`(5 sqrt 3-5

Solution:

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
Q 2807412388

The shadow of a pole standing on a horizontal plane is `d` metre longer when the Sun's altitude is a, then when it is `beta`· What is the height of the pole?

(A)

` d . ( cos alpha cos beta)/(cos (alpha - beta))`

(B)

` d . ( sin alpha cos beta)/(sin (alpha - beta))`

(C)

` d . ( sin alpha sin beta)/(sin ( beta - alpha))`

(D)

` d . ( sin beta sin alpha)/(sin ( alpha - beta))`

Solution:

Let CD be the pole, whose length is b.

In `Delta BCD, tan beta = (CD)/(BC) => tan beta = b/x`

`=> x = b cot beta` ... (i)

and in`Delta ACD, tan alpha = (CD)/(AC)`

`=> tan alpha = b/(x + d) => x + d = b cot alpha`

`=> b cot beta + d = b cot alpha ` [from Eq. (i)]

`=> b (cot alpha - cot beta )= d`

` :. b ( d sin alpha sin beta)/( cos alpha sin beta - cos beta sin alpha)`

`= (d sin alpha sin beta)/(sin (alpha - beta))`
Correct Answer is `=>` (C) ` d . ( sin alpha sin beta)/(sin ( beta - alpha))`
Q 2817412380

The angles of elevation of the top of a tower from two points at distances a and b metres from the
base and in the same straight line with it, are complementary. The height of the tower (in metres) is

(A)

`sqrt( a + b)`

(B)

`sqrt( a - b)`

(C)

`sqrt(ab)`

(D)

`sqrt(a/b)`

Solution:

Let `AB` be the tower and `C` and `D` be

the points of observation.

Then, `AC = a, AD = b,`

`angle ACB = theta , angle ADB = 90° - theta`

`(AB)/(AC) = tan theta`

`=> b = a tan theta`

and `(AB)/(AD) = tan (90^o - theta ) =cot theta`

`=> b = b cot theta , b^2 = ab`

Hence, `b = sqrt(ab)`
Correct Answer is `=>` (C) `sqrt(ab)`
Q 2817101989

A chimney `20` m high standing on the top of a building subtends an angle whose tangent is `116` at a distance `70` m from the foot of the building. The height of the building is

(A)

`50 m`

(B)

`40 m`

(C)

`60 m`

(D)

`20 m`

Solution:

Let `AB = H` m be the building and

`CB` be the chimney.

`tan theta = 1/6 , tan alpha = (H + 20)/(70)`

`tan beta = H/(70)`

Now, `tan theta = 1/6 => tan (alpha - beta) = 1/6`

`=> ( (tan alpha - tan beta))/( 1 + tan alpha tan beta) = 1/6`

`=> 6 ( (H + 20)/(70) - H/(70) )`

`= 1 + ( (H + 20)/(70) . H/(70))`

` => 6 xx (20)/(70) = ( 4900 + 20 H + H^2)/(70 xx 70)`

`=> H^2 + 20H + 4900 - 8400 = 0`

`:. H^2 + 20 H - 3500 = 0`

`=> (H - 50) (H + 70) = 0`

`:. H = 50m`
Correct Answer is `=>` (A) `50 m`

Calculating height and distance : Depression angel is given

Q 2713680540

From the top of a lighthouse, `100 m` high, the angle of depression of a boat is `tan^(-1) (5/12)`. What is the distance between the boat arid the light house?
NDA Paper 1 2017
(A)

`120 m`

(B)

`180 m`

(C)

`240 m`

(D)

`360 m`

Solution:

`tan theta = 100 /x`

`theta = tan ^-1\ \100/x`

`tan ^-1 \ \5 /12 = tan ^-1\ \ 100/x`

`x = 240 m`
Correct Answer is `=>` (C) `240 m`
Q 2711580420

A moving boat is observed from the top of cliff of 150 m height . The angle of depression of the boat changes from 60° to 45° in 2 minutes. What is the speed of the boat in metres per hour ?
NDA Paper 1 2016
(A)

`4500/sqrt3`

(B)

`(4500(sqrt3 -1))/sqrt3`

(C)

`4500sqrt3`

(D)

`(4500(sqrt3 + 1))/sqrt3`

Solution:

Let speed of boat per hr`= v`

Distance covered in 2 min = `( v * 2/60)`

In meter per hour `=v/30`

In `Delta ABC`

`tan 60= 150/x`

`x= 150/(sqrt 3)`

In `Delta ABD`

`tan 45= (x+ v/30)/150`

put the value of `x`

`150 = 150/(sqrt 3) + v/30`

`v= (4500 (sqrt 3-1))/(sqrt 3)`
Correct Answer is `=>` (B) `(4500(sqrt3 -1))/sqrt3`
Q 1772634536

From an aeroplane above a straight road the anlges of
depression of two positions at a distance `20 m` apart on
the road are observed to be `30^0` and `45^0`. The height of
the aeroplane above the ground is
NDA Paper 1 2014
(A)

`10sqrt(3) m`

(B)

`10 (sqrt(3) - 1) m`

(C)

`10 (sqrt(3) + 1) m`

(D)

`20 m`

Solution:

Let the height of the aeroplane above the

ground is `h` and `QB = x m`.

Given that, `PQ = 20m , angel APB = 30^0` and `angle AQB = 45^0`

Now, in `Delta AQB`,

` tan 45^0 = (AB)/(QB) = h/x`

` => 1 = h/x`

` => x = h` .........(1)

and in `Delta APB`

` tan 30^0 = (AB)/(PB) = h/(PQ + QB)`

` => 1/sqrt(3) = h/(20 + x) `

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

` => (sqrt(3) - h) = 20 + x = 20 + h`[from Eq. (i)]

` => (sqrt(3) h - h ) = 20`

` => h (sqrt(3) -1) = 20`

` :. h = (20)/(sqrt(3) - 1) . (sqrt(3) + 1)/(sqrt(3) - 1)`

` = (20 (sqrt(3) + 1))/(3-1) = (20 (sqrt(3) + 1))/2`

` = 10 (sqrt(3) + 1)`

Hence, the required height is `10 (sqrt(3) + 1) m`.
Correct Answer is `=>` (C) `10 (sqrt(3) + 1) m`
Q 2319667510

From the top of a lighthouse `70 m` high with its
base at sea level, the angle of depression of a boat
is `15°`. The distance of the boat from the foot of
the lighthouse is
NDA Paper 1 2012
(A)

`70 (2-sqrt 3)m`

(B)

`70 (2+sqrt 3)m`

(C)

`70(3-sqrt 3)m`

(D)

`70(3+sqrt 3)m`

Solution:

In `Delta ACB, tan15°=(AB)/(BC)`

`=> tan (45^o-30^o) =70/x`

`=>(1-1/sqrt 3)/(1+1/sqrt 3)=70/x=> (sqrt 3-1)/(1+sqrt 3)=70/x`

`= (70(3+1+2 sqrt 3))/(3-1)`

`=(70(4+2 sqrt 3))/2`

`70(2+sqrt 3)`

Hence, the required distance is `70 (2 + sqrt 3) m.`
Correct Answer is `=>` (B) `70 (2+sqrt 3)m`
Q 2339767612

From the top of a building of height `h m`, the angle
of depression of an object on the ground is `theta`.
What is the distance (in m) of the object from tl'\e
foot of the building?
NDA Paper 1 2012
(A)

`h cot theta`

(B)

`h tan theta`

(C)

`h cos theta`

(D)

`h sin theta`

Solution:

In `Delta BAC`

`tan theta=h/x`

`:. x=h cot theta`

So, the required distance is `h cot theta`.
Correct Answer is `=>` (A) `h cot theta`
Q 2319178019

From the top of a lighthouse `120 m` above the sea,
the angle of depression of a boat is `15^o`. What is
the distance of the boat from the lighthouse'?
NDA Paper 1 2010
(A)

`400 m`

(B)

`421 m`

(C)

`444 m`

(D)

`460 m`

Solution:

In `Delta ACB`

`tan 15^o=(AB)/(BC)=tan (45^o-30^o)`

`=> (tan 45^o-tan 30^o)/(1+tan 45^o tan 30^o)=120/(BC)=(1-1/sqrt 3)/(1+1/sqrt 3)`

`=> (sqrt 3-1)/(sqrt 3+1)=120/(BC)`

`:. BC=120((sqrt 3+1)^2)/(3-1)=60(3+1+2 sqrt 3)`

`=60(4+2 xx 1.73)`

`=60 xx 7.46 =447.6 m=444m` (approx)
Correct Answer is `=>` (C) `444 m`
Q 2143545443

From a 60 meter high tower angles of depression of the top and bottom of a house are `alpha` and `beta` respectively. If the height of the house is `(60 sin (beta-alpha))/x`, then x =
JEE Mock Mains
(A)

`sin alpha sin beta`

(B)

`cos alpha cos beta`

(C)

`sin alpha cos beta`

(D)

`cos alpha sin beta`

Solution:

`H = d tan beta ` and ` H-h = d tan alpha`

`=> 60/(60-h) = (tan beta)/(tan alpha) => -h = (60 tan alpha -60 tan beta)/( tan beta)`

`=> h = ( 60 sin (beta -alpha))/(cos alpha cos beta (sin beta)/(cos beta)) => x = cos alpha sin beta`
Correct Answer is `=>` (D) `cos alpha sin beta`
Q 1500723618

From the top of a hill `h` metres high the angles of depressions of the top and the bottom of pillar are `alpha` and `beta` respectively. The height (in metres) of the pillar is
BITSAT 2008
(A)

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

(B)

`(h tan beta- tan alpha)/(tan^2 beta)`

(C)

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

(D)

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

Solution:

Let `AB` be a hill whose height is h metres and CD be a pillar of height `h’ ` metres.

In `triangle` EDB,

`tan alpha = (h-h^')/ED...... (1)`


and in `triangleACB` ,

`tan beta= h/(AC) = h/(ED) .....................(2)`

Eliminate ED from Eqs. (i) and (ii), we get

`tan alpha = (h-h^')/h/(tan beta)`

`h ((tan alpha)/(tan beta)) = h-h^'`

`h^'` = `(h (tan beta- tan alpha))/(tan beta)`
Q 2458145004

From the top of a cliff `50` m high, the angles of
depression of the top and bottom of a tower
are observed to be `30°` and `45°`. The height of
tower is
UPSEE 2009
(A)

`50 m`

(B)

`50 sqrt3 m`

(C)

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

(D)

`50 ( 1 - sqrt3/3 ) m`

Solution:

Let the height of the cliff be `BD = 50` m and

height of the tower be `AE = h` metre.

In `Delta DEC`,

`tan 30° = ( 50 - h)//x `

`=> x = ( 50 - h)/(1//sqrt3) = sqrt3 ( 50 - h)` ......(i)

and in ` Delta BAD`,

`tan 45° = (50)/x => x = 50` m

From Eq. (i),

` 50 = sqrt(3) (50 - h)`

` => h = (50(sqrt3 -1))/sqrt3 ` m

` => h = 50 (1 - sqrt3/3 ) m`
Correct Answer is `=>` (D) `50 ( 1 - sqrt3/3 ) m`

Calculation of speed or time of observer

Q 2711580420

A moving boat is observed from the top of cliff of 150 m height . The angle of depression of the boat changes from 60° to 45° in 2 minutes. What is the speed of the boat in metres per hour ?
NDA Paper 1 2016
(A)

`4500/sqrt3`

(B)

`(4500(sqrt3 -1))/sqrt3`

(C)

`4500sqrt3`

(D)

`(4500(sqrt3 + 1))/sqrt3`

Solution:

Let speed of boat per hr`= v`

Distance covered in 2 min = `( v * 2/60)`

In meter per hour `=v/30`

In `Delta ABC`

`tan 60= 150/x`

`x= 150/(sqrt 3)`

In `Delta ABD`

`tan 45= (x+ v/30)/150`

put the value of `x`

`150 = 150/(sqrt 3) + v/30`

`v= (4500 (sqrt 3-1))/(sqrt 3)`
Correct Answer is `=>` (B) `(4500(sqrt3 -1))/sqrt3`
Q 2857312284

A man on the top of a tower, standing on the sea-shore finds that a boat coming towards him takes `10` min for the angle of depression to change from `30°` to `60°`. The time taken by the boat to reach the shore from this position will be

(A)

`5` min

(B)

`15` min

(C)

`7 1/2` min

(D)

`245 s`

Solution:

Let AB be the tower and C and D be

the two positions of the boat.

Let `AB = b , CD = x` and `AD = y`.

In `Delta BAD , y/b = cot 60° = 1/sqrt3 => y = b/sqrt3`

In `Delta BAC, (x + y)/b = cot 30° = sqrt3`

` => x + y = sqrt 3 b`

`:. x = (x + y) - y = sqrt 3 b - b/sqrt3 = (2b)/sqrt3`

`(2b)/sqrt3` is covered in `10` min.

` b/sqrt3` will be covered in `(10 xx sqrt3/(2b) xx b/sqrt3 )`

`=5` min
Correct Answer is `=>` (A) `5` min
Q 1650267114

A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is ` 30 ^{ \circ `}. After walking for `10` minutes from `A` in the same direction, at a point `B`, he observes that the angle of elevation of the top of the pillar is ` 60 ^{ \circ }`. Then the time taken (in minutes) by him, from `B` to reach the pillar is:
JEE Mains 2016
(A)

`6`

(B)

`10`

(C)

`20`

(D)

`5`

Solution:

`\tan 60^o =\ frac{b}{y}=\sqrt3`

`\tan 30 ^o=\ frac{b}{x+y}`

`\tan 30^o =\ frac{1}{\sqrt3}`

Take `t_1 =10` min

`x+y=b\sqrt 3`

As `x =10 p `

`p` be the speed

`10p+pt=\sqrt 3 y \sqrt3 ` (Put `y=p t,x=10p,b=\sqrt 3 y`)

`\Rightarrow 10p+pt=3pt`

`\Rightarrow 10+t=3t`

`\Rightarrow 10 =2t`

` t=5`
Correct Answer is `=>` (D) `5`
 
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