Mathematics Must Do Problems of Height and Distance For NDA

Please Wait... While Loading Full Video

Must Do Problems of Height and Distance For NDA

Q 2867445385

AB is a vertical pole. The end A is on the ground, C is the middle point of AB and P is a point on the level ground. The portion BC subtends an angle `alpha` at P. If `AP = n. AB`, then `tan alpha` is

(A)

`n/(n^2 - 1)`

(B)

`n/(n^2 + 1)`

(C)

`n/(2n^2 + 1)`

(D)

`((n^2 - 1)/(n^2 + 1))`

Solution:

Let `angle APC = theta` Then,

`tan(theta + alpha) = (AB)/(AP) = 1/n`

`tan (theta + alpha) = (AC)/(AP) = 1/2 . (AB)/(AP) = 1/(2n)`

`tan (theta + alpha ) = (tan theta + tan alpha)/(1 - tan theta tan alpha)`

` => 1/n = ( 1/(2n) + tan alpha)/(1 - 1/(2n) tan alpha)`

` => tan alpha = n/(1 + 2n^2)`

Correct Answer is `=>` (C) `n/(2n^2 + 1)`

Q 2817645580

A round balloon of radius `r` subtends an angle `alpha` at the eye of the observer while the angle of elevation of its centre is `beta`. Then height of the centre of the balloon is
I. `h = r sin alpha quad beta` II. `h = r sin beta sin alpha`
III. `h = r cosec (alpha/2) sin beta`

(A)

Only I

(B)

Only II

(C)

Only III

(D)

None of these

Solution:

Let `C` be the centre of the balloon

and AC and BC be radii, where OA and

OB are tangents from O.

Draw `CD bot OD`.

`angle AOB = alpha , angle BOC = alpha/2`

and `angle COD = beta , (OC)/(AC) = cosec alpha/2`

`=> OC = r cosec alpha/2 , (CD)/(OC) = sin beta`

`=> CD= OC sin beta`

`= r sin beta cosec (alpha //2)`

Correct Answer is `=>` (C) Only III

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 2817001889

Two verticals poles AL and BM of heights `20` m and `80` m, respectively stand apart on a horizontal plane. If A and B be the feet of the poles and AM and BL intersect at P, the height of P is equal to

(A)

`50m`

(B)

`18m`

(C)

`16m`

(D)

`15m``

Solution:

Let b be the height of point P from

horizontal plane.

In `Delta ABM` and `Delta AQP`

`tan theta = (80)/(AB) = b/(QA)`

`=> AQ = b · (AB)/(80)`

Now, in `Delta BAL` and `Delta BQP`

`tan phi = (20)/(AB) = b/(BQ)`

`:. BQ = b· (AB)/(20)`

Now, `AB = AQ + BQ`

`= b · AB (1/(80) + 1/(20))`

` => 1 = b ((1 + 4)/(80)) => b = 16 m`

Correct Answer is `=>` (C) `16m`

Q 2867212185

Two poles of equal heights are standing opposite to each other on either side of a road, which is `30` m wide. From a point between them on the road, the angles of elevation of the tops are `30°` and `60°`. The height of each pole is

(A)

`(15)/2 sqrt3 m`

(B)

`15 sqrt3 m`

(C)

`10 sqrt3 m`

(D)

None of these

Solution:

Let `AB` and `CD` be the poles and let

`O` be the point of observation.

Let ` AB = CD = b` m

`angle AOB = 30° , angle COD = 60°`.

Let `OA = x`.

Then, `OC = (30 - x) m`

In `Delta ,x/b = cot 30° = sqrt3`

`=> x = b sqrt3` ... (i)

In `Delta DCO , (30 - x)/b = cot 60° = 1/sqrt3`

`=> x = ( 30 - b/sqrt3)`

`:. b sqrt 3 = 30 - b/sqrt3` [from Eq. (i)]

`=> ( b sqrt3 + b/sqrt3 ) = 30`

` => 4b = 30 sqrt 3 => b = (15)/2 sqrt3 m`

Correct Answer is `=>` (A) `(15)/2 sqrt3 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 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`

Q 2827645581

The angle of elevation of a cloud from a point `200` m above a lake is `30°` and the angle of depression of its reflection in the lake is `60°`.
The height of the cloud above the lake is

(A)

`240 m`

(B)

`500 m`

(C)

`300 m`

(D)

`400 m`

Solution:

Let `AB` be the upper layer

of the lake, `C` be the cloud, `C'` its

reflection and `O` be the point of

observation. Draw `OD bot CC'`. Then,

`CB = C' B = x, angle COD = 30°`,

`angle C'OD = 60°, AO = BD = 200 m`

`tan30° = (CD)/(OD)`

`=> 1/sqrt3 = (CB - BD)/(AB) = (x - 200)/(AB)`

`=> x - 200 = (AB)/sqrt3` ......(i)

`tan 60° = (C'D)/(OD)`

`=> sqrt3 = (C' B + BP)/(AB) = ( x + 200)/(AB)`

`=> x + 200 = AB sqrt3` ... (ii)

On dividing Eq. (i) by Eq. (ii), we get

`(x - 200)/(x + 200) = 1/3`

`=> 3x - 600 = x + 200`

`=> x = 400 m`

Height of the cloud above the lake

`= x = 400 m`

Correct Answer is `=>` (D) `400 m`

Q 2807545488

Let PT be 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 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
I. The value of `x` is `8`.
II. The value of `theta` is `pi/4`.
Which of the above statement(s) is/are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

`∵ PT = 2^x m , AP = 2^(x + 1) m`

and `BP =192 m`

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

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

Now, in `Delta PTB, tan 2 theta = (PT)/(PB) = 2^x/(192)`

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

`=> 4/3 xx 192 = 2^x => 2^x = 256`

` => 2^x = 2^8 => x = 8`

`tan theta = (TP)/(AP) = 2^8/2^9 = 1/2`

` theta = tan^(-1) (1/2) != pi/4`

Correct Answer is `=>` (A) Only I

Q 2847512483

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 `alpha` and `beta`, respectively and if `AB = d` m, what is `h//d` equal to?

(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 ADC, tan alpha = b/x => x = b cot alpha ` ... (i)

In ` Delta CDB, tan beta = b/( d - x)`

`=> d - x = b cot beta ` ... (ii)

On adding Eqs. (i) and (ii), we get

`d = b (cot alpha + cot beta )`

We know that,

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

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

`=> d = b [ (cot alpha cot beta -1)/(cot (alpha + beta))]`

` :. b/d = ( cot (alpha + beta))/( cot alpha cot beta - 1)`

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

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 2837645582

An aeroplane flying horizontally `900` m above the ground is observed at an elevation of `60°`. After `10 s`, the elevation changes to `30°`.
The uniform speed of the aeroplane (in km/h) is

(A)

`(50)/sqrt3`

(B)

`180 sqrt3`

(C)

`216 sqrt3`

(D)

None of these

Solution:

Let A and B be the two

positions of the aeroplane and `O` be the

point of observation.

`(OD)/(BD) = cot 30° = sqrt3 => OD = 900 sqrt3 m`

`(OC)/(AC) = cot 60° = 1/sqrt3 => OC = (900)/sqrt3 m`

`:. AB = CD = OD - OC`

`= ( 900 sqrt3 - (900)/sqrt3 ) = (1800)/sqrt3 m`

Speed = `text( Distance)/text(Time) = ( (1800)/(sqrt3 xx 10) )` m/s

`= ( (180)/sqrt3 xx (18)/5 = 216 sqrt3` km/h

Correct Answer is `=>` (C) `216 sqrt3`

Q 1711412329

A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is `45^{\circ}`. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to `30^{\circ}`. Then the speed (in m/s) of the bird is
JEE Mains 2014

(A)

`40(\sqrt{2}-1)`

(B)

`40(\sqrt{3}-\sqrt{2})`

(C)

`20\sqrt{2}`

(D)

`20(\sqrt{3}-1)`

Solution:

`\tan 30^{\circ}= \frac{20}{20+x}= \frac{1}{\sqrt{3}}`

`20+x=20\sqrt{3}`

`x=20(\sqrt{3}-1)`

`\Rightarrow` Speed is `20(\sqrt{3}-1)` m/sec

Correct Answer is `=>` (D) `20(\sqrt{3}-1)`