Heights And Distances Class 10th Mathematics Gujarat Board Solution

Class 10th Mathematics Gujarat Board Solution
Exercise 10
  1. A pole stands vertically on the ground. If the angle of elevation of the top of…
  2. A string of a kite is 100 m long and it makes an angle of measure 60 with the…
  3. A circus artist is climbing from the ground along a rope stretched from the top…
  4. A tree breaks due to a storm and the broken part bends such that the top of the…
  5. An electrician has to repair an electric fault on the pole of height 5 m. He…
  6. As observed from a fixed point on the bank of a river, the angle of elevation…
  7. As observed from the top of a hill 200 m high, the angles of depression of two…
  8. A person standing on the bank of a river, observes that the angle subtended by…
  9. The shadow of a tower is 27 m, when the angle of elevation of the sun has…
  10. From a point at the height 100 m above the sea level, the angles of depression…
  11. From the top of a 300 m high light - house, the angles of depression of the…
  12. As observed from a point 60 m above a lake, the angle of elevation of an…
  13. Watching from a window 40 m high of a multi - storeyed building, the angle of…
  14. Two pillars of equal height stand on either side of a road, which is 100 m…
  15. The angles of elevation of the top of a tower from two points at distance a…
  16. A man on the top of a vertical tower observes a car moving at a uniform speed…
  17. If the angle of elevation of a cloud from a point h metres above a lake has…
  18. From the top of a building vector ab , 60 m high, the angles of depression of…
  19. A bridge across a valley is h metres long. There is a temple in the valley…
  20. At a point on level ground, the angle of elevation of a vertical tower is…
  21. A statue 1.46 m tall, stands on the top of a pedestal. From the point on the…
  22. On walking ……….. metres on a hill making an angle of measure 30 with the…
  23. The angle of elevation of the top of the tower from a point P on the ground…
  24. A 3 m long ladder leans on the wall such that its lower end remains 1.5 m…
  25. A tower is 50√3 m high. The angle of elevation of its top from a point 50 m…
  26. If the ratio of the height of a tower and the length of its shadow is 1 : √3,…
  27. If the angles of elevation of a tower from two points distance a and b (a b)…
  28. The tops of two poles of height 18 m and 12 m are connected by a wire. If the…
  29. The angle of elevation of the top of the building A from the base of building…
  30. If the angle of elevation of the top of a tower of a distance 400 m from its…
  31. The angle of depression of a ship from the top of a tower 30 m height has…
  32. When the length of the shadow of the pole is equal to the height of the pole,…
  33. From the top of a building h metre high, the angle of depression of an object…
  34. As observed from the top of the light house the angle of depression of the…
  35. Two poles are x metres apart and the height of one is double than that of the…

Exercise 10
Question 1.

A pole stands vertically on the ground. If the angle of elevation of the top of the pole from a point 90 m away from the pole has measure 30, find the height of the pole.


Answer:

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.



Question 2.

A string of a kite is 100 m long and it makes an angle of measure 60 with the horizontal. Find the height of the kite, assuming that there is no slack in the string.


Answer:

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.



Question 3.

A circus artist is climbing from the ground along a rope stretched from the top of a vertical pole and tied at the ground. The height of pole is 10 m and the angle made by the rope with ground level has measure 30. Calculate the distance covered by the artist in climbing to the top of the pole.


Answer:

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.



Question 4.

A tree breaks due to a storm and the broken part bends such that the top of the tree touches the ground making an angle having measure 30 with the ground. The distance from the foot of the tree to the point where the top touches the ground is 30 m. Find the height of the tree.


Answer:

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.



Question 5.

An electrician has to repair an electric fault on the pole of height 5 m. He needs to reach a point 2 m below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use, which when inclined at an angle of measure 60 to the horizontal would enable him to reach the required position.


Answer:

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.



Question 6.

As observed from a fixed point on the bank of a river, the angle of elevation of a temple on the opposite bank has measure 30. If the height of the temple is 20 m, find the width of the river.


Answer:

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.



Question 7.

As observed from the top of a hill 200 m high, the angles of depression of two vehicles situated on the same side of the hill are found to have measure 30 and 60 respectively. Find the distance between the two vehicles.


Answer:

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.



Question 8.

A person standing on the bank of a river, observes that the angle subtended by a tree on the opposite bank has measure 60. When he retreats 20 m from the bank, he finds the angle to have measure 30. Find the height of the tree and the breadth of the river.


Answer: 


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



Question 9.

The shadow of a tower is 27 m, when the angle of elevation of the sun has measure 30. When the angle of elevation of the sun has measure 60, find the length of the shadow of the tower.


Answer:

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.



Question 10.

From a point at the height 100 m above the sea level, the angles of depression of a ship in the sea is found to have measure 30. After some time the angle of depression of the ship has measure 45. Find the distance travelled by the ship during that time interval.


Answer:

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.



Question 11.

From the top of a 300 m high light – house, the angles of depression of the top and foot of a tower have measure 30 and 60. Find the height of the tower.


Answer:

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.



Question 12.

As observed from a point 60 m above a lake, the angle of elevation of an advertising balloon has measure 30 and from the same point the angle of depression of the image of the balloon in the lake has measure 60. Calculate the height of the balloon above the lake.


Answer:

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.



Question 13.

Watching from a window 40 m high of a multi – storeyed building, the angle of elevation of the top of a tower is found to have measure 45. The angle of elevation of the top of the same tower from the bottom of the building is found to have measure 60. Find the height of the tower.


Answer:

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.



Question 14.

Two pillars of equal height stand on either side of a road, which is 100 m wide. The angles of elevation of the top of the pillars have measure 60 and 30 at a point on the road between the pillars. Find the position of the point from the nearest end of a pillars and the height of pillars.


Answer:

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.



Question 15.

The angles of elevation of the top of a tower from two points at distance a and b metres from the base and in the same straight line with it are complementary. Prove that the height of the tower is  metres.


Answer:

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.



Question 16.

A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change its measure from 30 to 45, how soon after this, will the car reach the tower?


Answer:

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.



Question 17.

If the angle of elevation of a cloud from a point h metres above a lake has measure a and the angle of depression of its reflection in the lake has measure β, prove that the height of the h(tani3 + tana) cloud is  .


Answer:

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.



Question 18.

From the top of a building , 60 m high, the angles of depression of the top and bottom at a vertical lamp post  are observed to have measure 30 and 60 respectively. Find,

(1) the horizontal distance between building and lamp post.

(2) the height of the lamp post.

(3) the difference between the heights of the building and the lamp post.


Answer:

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



Question 19.

A bridge across a valley is h metres long. There is a temple in the valley directly below the bridge. The angles of depression of the top of the temple from the two ends of the bridge have measures α and β. Prove that the height of the bridge above the top of the temple is 


Answer:

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 .



Question 20.

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is . On walking 192 metres towards the tower, the tangent of the angle is found to be. Find the height of the tower.


Answer:


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.



Question 21.

A statue 1.46 m tall, stands on the top of a pedestal. From the point on the ground the angle of elevation of the top of the statue has measure 60 and from the same point, the angle of elevation of the top of the pedestal has measure 45. Find the height of the pedestal.


Answer:

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.



Question 22.

On walking ……….. metres on a hill making an angle of measure 30 with the ground, one can reach the height of 'a' metres from the ground.
A. 

B. 

C. 2a

D. 


Answer:


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




H = 2a


Question 23.

The angle of elevation of the top of the tower from a point P on the ground has measure 45. The distance of the tower from the point P is a and height of the tower is b. Then, ……….
A. a > b

B. a ∠ b

C. a = b

D. a = 2b


Answer:


From the figure we see that,




A = b


Question 24.

A 3 m long ladder leans on the wall such that its lower end remains 1.5 m away from the base of the wall. Then, the ladder makes an angle of measure ……… with the ground.
A. 30

B. 45

C. 60

D. 20


Answer:


From the figure we see that,




But, 


θ = 60⁰


Question 25.

A tower is 50√3 m high. The angle of elevation of its top from a point 50 m away from its foot has measure
A. 45

B. 60

C. 30

D. 15


Answer:


From the figure we see that,



Tan θ = √3


But, tan 60 = √3


θ = 60⁰


Question 26.

If the ratio of the height of a tower and the length of its shadow is 1 : √3, then the angle of elevation of the sun has measure ………….
A. 30

B. 45

C. 60

D. 75


Answer:


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


Now, 


From the figure we see that,



But 


θ = 30⁰


Question 27.

If the angles of elevation of a tower from two points distance a and b (a > b) from its foot on the same side of the tower have measure 30 and 60, then the height of the tower is ……….
A. 

B. 

C. 

D. 


Answer:

Let the height of the tower be h.



From the figure we see that,


 and 


Now, 



H2 = ab


H = √ab


Question 28.

The tops of two poles of height 18 m and 12 m are connected by a wire. If the wire makes an angle of measure 30 with horizontal, then the length of the wire is
A. 12 m

B. 10 m

C. 8 m

D. 4 m


Answer:

Let the length of the wire be l.



From the figure we see that,




L = 12 m


Question 29.

The angle of elevation of the top of the building A from the base of building B has measure 50. The angle of elevation of the top of the building B from the base of building A has measure 70. Then, ………….
A. building A is taller than building B.

B. Building B is taller than building A.

C. Building A and building B are equally tall.

D. The relation about the heights of A and B cannot be determined.


Answer:

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


Question 30.

If the angle of elevation of the top of a tower of a distance 400 m from its foot has measure 30, then the height of the tower is ……..
A. 200√2

B. 

C. 

D. 


Answer:


From the figure we see that,



H = 400 tan 30




Question 31.

The angle of depression of a ship from the top of a tower 30 m height has measure 60. Then, the distance of the ship from the base of the tower is ………..
A. 10

B. 30

C. 10√3

D. 30√3


Answer:


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


From the figure we see that,





h = 10√3


Question 32.

When the length of the shadow of the pole is equal to the height of the pole, then the angle of elevation of the source of light has measure
A. 45

B. 30

C. 60

D. 75


Answer:


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


Question 33.

From the top of a building h metre high, the angle of depression of an object on the ground has measure θ. The distance (in metres) of the object from the foot of the building is
A. h sinθ

B. h tanθ

C. h cotθ

D. h cosθ


Answer:


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


Question 34.

As observed from the top of the light house the angle of depression of the two ships P and Q anchored in the sea to the same side are found to have measure 35 and 50 respectively. Then from the light house....
A. P and Q are at equal distance.

B. The distance of Q is more then P.

C. The distance of P is more than Q.

D. The relation about the distance of P and Q cannot be determined.


Answer:


From the figure we see that,


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


Question 35.

Two poles are x metres apart and the height of one is double than that of the other. If from the mid – point of the line joining their feet, an observer finds the angle of elevation of their tops to be complementary, then the height of the shorter pole is
A. 

B. 

C. 

D. 


Answer:

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  .


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