Heights And Distances

Suppose AB represents the pole. C is a point 90m away from the pole.

AB = Height of the pole

BC = 90m

∠B is a right angle in ∆ABC and ∠C = 30⁰

In ∆ABC,

Thus, the height of the pole is 51.9 m.

Let PR be the string of a kite P and QR be the horizontal direction.

PR = 100 m

∠PQR = 90⁰

∠PRQ = 60⁰

PQ = Height of the kite

In ∆PQR,

PQ = 50√3

= 50 × 1.73

= 86.5 m

The height of a kite is 86.5 m.

Let PQ be the pole, whose height is 10 m and PR be the length of a rope stretched from the top of a vertical pole. The artist is moving from R to P.

The angle made by the rope with ground level has measure 30⁰.

∠PRQ = 30⁰

In ∆PQR,

PR = 10 × 2

= 20 m

The distance covered by the artist in climbing to the top of the pole is 20m.

Suppose AC is the tree broken at point B such that the broken part CB takes the position BD and touches the ground at D.

AD = 30 m

∠ADB = 30⁰

In ∆DAB,

AB = 10√3

Now,

BD = 20√3

So, height of the tree AC,

= AB + BC

= AB + BD

= 10√3 + 20√3

= 30√3

= 30 × 1.73

= 51.9 m

The height of the tree is 51.9 m.

Let PR represent a polle oof length 5 m.

Q is a point 2 m below the top of the pole P to undertake the repair work.

PQ = 2 m

QR = 3 m

QS = Length of the ladder

∠QSR = 60⁰

In ∆QRS,

QS = 2√3

= 2 × 1.73

= 3.46 m

The length of ladder is 3.46 m.

Let PQ represent the temple and P is the top of the temple.

PQ = 20 m

R is a fixed point on the bank of a river from a temple on the opposite bank and QR is the width of the river.

∠PRQ = 30⁰

In ∆PQR,

QR = 20√3

= 20 × 1.73

= 34.6 m

The width of the river is 34.6 m.

Let PQ be the hill with height = 200 m

R and S are two vehicles situated on the same side of the hill.

The angles of depression of vehicles R and S from P are 60⁰ and 30⁰ respectively, so

∠PRQ = ∠XPR = 60⁰

∠PSQ = ∠XPS = 30⁰

SR = Distance between the two vehicles

In ∆PQR,

In ∆PQS,

QS = 200√3

Distance between the two vehicles situated on the same side of the hill = SR

= QS – QR

= 230.6

The distance between two vehicles is 230.36 m.

Let PM be the tree. OM represent the width of the river.

So, ∠PMO = 90⁰ and OA = 20 m

Let OM = Breadth of the river = x and

PM = Height of a tree = h

AM = OM + OA = x + 20

In ∆PMO,

H = √3x …………(1)

In ∆PAM,

……………(2)

From (1) and (2) we get,

X + 20 = √3x × √3

X + 20 = 3x

3x – x = 20

2x = 20

X = 10

Now, h = √3x

= 1.73 × 10

= 17.3

The breadth of the river = OM = x = 10 m and

The height of the tree = PM = h = 17.3 m

Let AB be the tower and CB its shadow when the angle of elevation of the sun is 30⁰.

∠ACB = 30⁰

∠B = 90⁰

CB = 27 m

DB is the shadow of the tower when the angle of elevation of the sun is 60⁰.

In ∆ACB,

In ∆ADB,

The length of the shadow of the tower is 9 m.

Let the ship travel from A to Bin given time. So, the distance travelled by ship is AB.

Suppose, the observation point is at O.

OC = 100 m

The angle of depressions of A and B from O are 30⁰ and 45⁰ respectively.

∠OAC = ∠XOA = 30⁰

∠OBC = ∠XOB = 45⁰

In ∆OCB,

BC = 100

In ∆OCA,

100 + AB = 173

AB = 173 – 100

= 73 m

The distance travelled by ship during the given time is 73 m.

Let AC be the lighthouse and ED be the tower.

Height of lighthouse = AC = 300 m

Let ED = h

Let EB be the perpendicular from E to AC.

The angles of depression of the top E and bottom D of the tower ED measures 30⁰ and 60⁰ respectively from A.

∠AEB = 30⁰

∠ADC = 60⁰

In ∆ADC,

Now, In ∆AEB,

AB = 100 m

Height of the tower ED = BC

= AC – AB

= 300 – 100

= 200 m

The height of the tower is 200m.

Let BE be the surface of the lake.

A is a point 60 m above a lake.

F is the image of balloon C in the lake.

Horizontal line AD intersects CE in D.

∠CAD = 30⁰, ∠FAD = 60⁰, AB = 60 m

Let CE = h, BE = l

Then CD = h – 60, DF = h + 60

In ∆ADC,

L = √3(h – 60)……….(1)

In ∆ADF,

√3l = h + 60…………(2)

From equation (1) and (2),

√3[√3(h – 60)] = h + 60

3(h – 60) = h + 60

3h – 180 = h + 60

2h = 240

H = 120 m

The height of balloon above the lake is 120 m.

Let CD be the window, AB be the tower and D the point of observation.

CD = 40 m

Let the height of the tower = AB = h

Horizontal line DE intersects AB in E.

BE = CD = 40 m

AE = AB – BE = (h – 40)

∠AED = ∠ABC = 90⁰

Now, the angle of elevation of A from D is 45⁰ and the angle of elevation of A from C is 60⁰.

∠ADE = 45⁰, ∠ACB = 60⁰

In ∆AED and in ∆AEC,

DE = h – 40 = BC ………..(1)

H = (h – 40)√3

H = √3h – 40√3

H(√3 – 1) = 40√3

= 60 + 20(1.73)

= 60 + 34.6

= 94.6 m

The height of tower is 94.6 m.

Let AB and DE be two pillars of equal height.

AB = DE

The angle of elevation of A and E from C are respectively 60⁰ and 30⁰ respectively.

∠ACB = 60⁰, ∠EDC = 30⁰ and BD = 100 m

Let BC = x

CD = BD – BC = 100 – BC = (100 – x)

In ∆ABC,

… …. (1)

In ∆EDC,

(AB = DE)

100 – x = √3AB

100√3 – AB = 3AB

4AB = 100√3

AB = 25√3

= 25 × 1.73

= 43.25 m

The height of the pillar is 43.25 m.

From equation(1),

= 25

The distance of the point from the nearest end of the pillars is 25 m and the height of each pillar is 43.25 m.

AB is a tower. D and Care two points on the same side of a tower, BD = a and BC = b.

∠ADB and ∠ACB are the complementary angles.

If ∠ADB = x, then ∠ACB = 90 – x

In ∆ADB,

………… (1)

In ∆ABC,

…………....(2)

Multiplying (1) and (2),

(AB)² = ab

AB = √ab

Height of tower = AB = √ab

Hence proved.

Let AB be the tower with height of tower AB = h

At C the angle of depression of car measures 30 and 12 minutes later it reaches D where angle of depression is 45.

Let CD = x, D = y

Here, AB = h, ∠ACB = ∠XAC = 30⁰

∠ADB = ∠XAD = 45⁰

In ∆ACB,

x + y = √3h ……(1)

In ∆ABD,

H = y ………(2)

From equation 1 and 2,

x + y = √3y

x = (√3 – 1)y

The distance covered by car in 12 minutes is CD = x

So, time taken to cover distance x is 12 minutes.

So, y = Time taken to cover distance DB

= 6 × 2.73

= 16.38 minutes

The time taken by car to reach the tower is 16.38 minutes.

Let AB be the surface of the lake. E is the point above h metre from A.

AE = h

Let height of the cloud BD = l

Let F be the reflection of kite C.

Horizontal line EC intersects BD in C.

BF = l

∠DEC = α, ∠CEF = ß

Here, AE = BC = h

CD = BD – BC = l – h and

CF = BF + BC = l + h

In ∆ECD,

……(1)

In ∆ECF,

…….(2)

From results (1) and (2),

Using componendo – dividendo,

The height of the cloud from the surface of the lake is m.

Hence proved.

Let AB be the building and CD be the lamppost.

The height of the building AB = 60 m

Horizontal line DE intersects AB in E.

Let BE = CD = x

AE = AB – BE = (60 – x) m

∠AED = ∠ABC = 90⁰

Now, the angle of depression of the top D and then bottom C of the post CD are 30⁰ and 60⁰ respectively from A.

Then, ∠ADE = ∠XAD = 30⁰ and

∠ACB = ∠XAC = 60⁰

In ∆ADE,

DE = √3(60 – x) ………(1)

In ∆ABC,

…………..(2)

Now, BC = DE

From (1) and (2),

√3(60 – x) = 20√3

60 – x = 20

X = 40

1) The horizontal distance between the building and the lamppost

= BC

= √3(60 – x)

= √3(60 – 40)

= 20√3

= 20 × 1.73

= 34.6

2) The height of the lamppost

= CD

= x

= 40 m

3) The difference between the heights of the building and the lamppost

= AB – BE

= 60 – x

= 60 – 40

= 20 m

Let AD be the bridge and E be the top of the temple.

The perpendicular from E on AD is EF.

Thus, EF represents the height from the top of the temple to the bridge.

Let EF = x and DF = y

AF = h – y

In ∆ABE,

………(1)

In ∆DCE,

………..(2)

From (1) and (2),

h tan α tan ß – x tan ß = x tan α

h tan α tan ß = x(tan α + tan ß)

The height of the bridge above the top of the temple is .

Let AB be the tower.

The angle of elevation of tower from D is α and on walking 192 metres towards the tower from D to B, the angle of elevation of tower from C is ß.

CD = 192 m

,

Let BC = x, AB = h

BD = BC + CD = x + 192

∠ADB = α ∠ACB = ß

In ∆ABC,

4h = 3x

…..equation (1)

In ∆ABD,

12h = 5x + 960

36h = 20h + 2880

16h = 2880

H = 180

The height of the tower = AB = h = 180 m.

Let AB be the pedestal and AB = x

BC is a statue and BC = 1.46 m

The angle of elevation of the top of the statue C from D is 60⁰ and from the same point the angle of elevation of the top of the pedestal B is 45⁰.

In ∆BAD,

AD = x

In ∆CAD,

√3x = x + 1.46

(√3 – 1)x = 1.46

(1.73 – 1)x = 1.46

0.73x = 1.46

X = 2

The height of the pedestal is = AB = x = 2 m.

Suppose on walking h metres, one can reach the height of “a” metres from the ground.

H = 2a

From the figure we see that,

A = b

From the figure we see that,

But,

θ = 60⁰

From the figure we see that,

Tan θ = √3

But, tan 60 = √3

θ = 60⁰

Let the height of tower be h and length of the shadow be a

Now,

From the figure we see that,

But

θ = 30⁰

Let the height of the tower be h.

From the figure we see that,

and

Now,

H² = ab

H = √ab

Let the length of the wire be l.

From the figure we see that,

L = 12 m

Let the height of the building B be b and the height of the building A be a and the distance between the two building be x.

From the figure we see that,

and

and

b>a

Height of building B> Height of building A

From the figure we see that,

H = 400 tan 30

Let the distance of the ship from the base of the tower be h.

From the figure we see that,

h = 10√3

Let the length of the pole be h and so the length of the shadow become h.

Suppose the angle of elevation of the sun is θ.

From the figure we see that,

θ = 45⁰

Let the distance of the object from the foot of the building be l.

From the figure we see that,

The distance of a point from the building l = h cot θ

From the figure we see that,

The distance of P from the lighthouse is more than the distance of Q from the lighthouse.

Let the height of the poles be h and 2h respectively.

Let the angles of elevation of the top of the poles from the midpoint M of the line joining the feet of the poles be α and ß.

Also, given that they are complementary angles, α + ß = 90⁰

From the figure we see that,

and

and

Now,

tan (90 – α) = 2 tan α

cot α = 2 tan α

Now,

The height of the shorter pole is .