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Page No 572:
Question 1:
A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60°. Find the height of the tower.
Answer:
Let be the tower standing vertically on the ground and O be the position of the observer.
We now have:
∠ and ∠
Let:
m
Now, in the right ∆, we have:
=
⇒
⇒ =
Hence, the height of the pole is 34.64 m.
Page No 572:
Question 2:
A kite is flying at a height of 75 m from the level ground, attached to a string inclined at a 60° to the horizontal. Find the length of the string assuming that there is no slack in it.
Answer:
Let be the horizontal ground and be the position of the kite.
Also, let O be the position of the observer and be the thread.
Now, draw ⊥ .
We have:
∠, m and ∠
Height of the kite from the ground = = 75 m
Length of the string, m
In the right ∆, we have:
⇒
⇒ m
Hence, the length of the string is m.
Page No 573:
Question 3:
An observe, 1.6 m tall, is 45 metres away from a tower. The angle of elevation from his eye to the top of the tower is 30°. Determine the height of the tower.
Answer:
Let be the horizontal ground and be the position of the observer.
Let be the position of the tower such that m.
Let = m be the height of the tower.
Now,
Height of the observer, = m
From the right ∆, we have:
⇒
⇒ m
∴ m [∵ CX = BY]
⇒ m
Hence, the height of the tower is m.
Page No 573:
Question 4:
The height of a tree is 10 m. It is bent by the wind in such a way that its top touches the ground and makes an angle of 60° with the ground. At what height from the bottom did the tree get bent?
Answer:
Let be the tree broken at point such that takes the position and ∠.
If m, then m.
In ∆, we have:
⇒
⇒
⇒ m
Now,
m
Hence, the height from the bottom from where the tree got bent is 4.64 m.
Page No 573:
Question 5:
A 1.5-m-tall boy stands at a distance of 3 m from a lamp post and casts a shadow of 4.5 m on the ground. Find the height of the lamp post.
Answer:
Let be the lamppost, be the boy and be the shadow of .
Also, let ∠ and the height of the lamppost be h.
From the right ∆, we have:
Now, from the right ∆, we have:
[∵ ]
⇒
⇒ m
∴ Height of the lamppost = m
Page No 573:
Question 6:
The angle of elevation of the top of a building from a point A on the ground is 30°. On moving a distance of 30 m towards its base to a point B, the angle of elevation changes to 45°. Find the height of the building and the distance of its base from the point A.
Answer:
Let be the building. Then, the angle of elevation from the top of the building on the ground is such that ∠.
After moving to point , the angle of elevation changes such that ∠ and m.
Let:
m and m
In the right ∆, we have:
⇒
Now, in the right ∆, we have:
On putting in the above equation, we get:
⇒
⇒
⇒ m
Hence, the height of the building is 40.98 m.
Now,
m
∴ Distance of the base of the building from point m
Page No 573:
Question 7:
The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 m towards the foot of the tower to a point B, the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the point A.
Answer:
Let be the tower making an angle of elevation at point on the ground such that ∠.
On moving towards the foot of the tower at point , m and ∠.
Let:
m and m
In the right ∆, we have:
⇒ ...(i)
Now, in the right ∆, we have:
⇒
⇒ ...(ii)
By using (ii) in (i), we get:
⇒
⇒
⇒ m
∴ Height of the tower = 17.32 m
Now,
m
∴ Distance of the tower from point = m
Page No 573:
Question 8:
From the top of a building 15 m high, the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60°. Find height of the tower, and the distance between the tower and the building.
Answer:
Let be the tower and be the building.
We have:
m, ∠ and ∠
Let m and m such that m.
In the right ∆, we have:
⇒
Now, in the right ∆, we have:
On putting in the above equation, we get:
⇒
⇒
⇒
⇒ m
Hence, the height of the tower is 22.5 m.
Now,
Distance between the tower and the building = m
Page No 573:
Question 9:
From a window 15 m high above the ground, in a street, the angles of elevation and depression of the top and foot of another house on the opposite side of the street are 30° and 45° respectively. Show that the height of the opposite house is 23.66 m.
Answer:
Let be the house and be the window.
We have:
m such that m, ∠ and ∠.
Let m such that m and let m such that m.
In the right ∆, we have:
⇒
⇒ m
Now, in the right ∆, we have:
⇒
On putting in the above equation, we get:
⇒
⇒
⇒
⇒ m
Hence proved.
Page No 573:
Question 10:
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff. At a point on the plane, 30 metres away from the tower, an observe notices that the angles of elevation of the top and bottom of the flagstaff are 60° and 45° respectively. Find the height of the flagstaff and that of the tower.
Answer:
Let be the horizontal line, be the vertical tower and be the vertical flagstaff.
We now have:
m, ∠ and ∠
Let:
m and m
In the right ∆, we have:
⇒
⇒ m
Now, in the right ∆, we have:
⇒
On putting in the above equation, we get:
⇒ m
We now have:
Height of the flagstaff = x = 21.96 m
Height of the tower = m
Page No 573:
Question 11:
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5 m. From a point on the plane the angles of elevation of the bottom and the top of the flagstaff are 30° and 60°. Find the height of the tower.
Answer:
Let be the horizontal line, be the vertical tower and be the vertical flagstaff such that m.
Let:
m and m
In the right ∆, we have:
⇒
⇒
Now, in the right ∆, we have:
⇒
On putting in the above equation, we get:
⇒
⇒
⇒
⇒ m
Hence, the height of the tower is 2.5 m.
Page No 573:
Question 12:
A statue 1.46 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Answer:
Let be the pedestal and be the statue such that m.
We have:
∠ and ∠
Let:
m and m
In the right ∆, we have:
⇒
⇒
Or,
Now, in the right ∆, we have:
⇒
On putting in the above equation, we get:
⇒
⇒
⇒ m m
Hence, the height of the pedestal is 2 m.
Page No 574:
Question 13:
The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30° On advancing 150 m towards the foot of the tower, the angle of elevation becomes 60°. Show that the height of the tower is 129.9 metres.
Answer:
Let be the tower.
We have:
m, ∠ and ∠
Let:
m and m
In the right ∆, we have:
⇒
⇒
Now, in the right ∆, we have:
⇒
⇒
On putting in the above equation, we get:
⇒
⇒
⇒ m
Hence, the height of the tower is 129.9 m.
Page No 574:
Question 14:
On a horizontal plane there is a vertical tower with a flagpole on the top of the tower. At a point, 9 metres away from the foot of the tower, the angle of elevation of the top of bottom of the flagpole are 60° and 30° respectively. Find the height of the tower and the flagpole mounted on it.
Answer:
Let be the horizontal plane, be the tower and be the vertical flagpole.
We have:
m, ∠ and ∠
Let:
m and m
In the right ∆, we have:
⇒
⇒ m
Now, in the right ∆, we have:
⇒
⇒
By putting in the above equation, we get:
⇒
⇒
Thus, we have:
Height of the flagpole = 10.39 m
Height of the tower = 5.19 m
Page No 574:
Question 15:
From the top of a hill 200 m high, the angles of depression of the top and bottom of a pillar are 30° and 60° respectively. Find the height of the pillar and its distance from the hill.
Answer:
Let be the hill and be the pillar. Draw ⊥ .
Thus, we have:
m, ∠ and ∠
Now, let m such that m and let m such that m.
In the right ∆, we have:
⇒
⇒ m
Now, in the right ∆, we have:
⇒
By putting in the above equation, we get:
⇒
⇒
⇒ m
We now have:
Height of the pillar = 133.33 m
Distance of the pillar from the hill = 115.47 m
Page No 574:
Question 16:
From a window, 60 metres high above the ground, of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are 60° and 45° respectively. Show that the height of the opposite house is 60 metres.
Answer:
Let be the house and be the window such that m. Draw ⊥ .
Thus, we have:
∠ and ∠
Let m and m such that m and m.
In the right ∆, we have:
⇒
⇒ m
Now, in the right ∆, we have:
⇒
On putting in the above equation, we get:
⇒
⇒ m
Hence, the height of the opposite house is m.
Page No 574:
Question 17:
The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of , find the speed of the jet plane.
Answer:
Let C and C' be the two positions of the jet plane and BC and B'C' be its constant heights.
Thus, we have:
Now, in the right ∆ABC, we have:
Now, in the right ∆AB'C', we have:
BB' is 3000 m. This means that to cover a distance of 3000 m, the jet plane takes 15 seconds.
∴ Speed of the jet plane
Page No 574:
Question 18:
The angles of elevation and depression of the top and bottom of a lighthouse form the top of a building, 60 m high, are 30° and 60° respectively. Find (i) the difference between the heights of the lighthouse and the building, and (ii) the distance between the lighthouse and the building.
Answer:
Let be the lighthouse and be the building. Draw ∥ .
We have:
m, ∠ and ∠
Let m such that m and m.
Let BC = x.
In the right ∆, we have:
⇒
⇒ m
Now, in the right ∆, we have:
⇒
On putting in the above equation, we get:
⇒
∴ Difference between the heights of the lighthouse and building = m
Distance between the lighthouse and the building = m
Page No 574:
Question 19:
As observed form the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation.
Answer:
Let be the lighthouse and B and C be the two positions of the ship.
Thus, we have:
m, ∠ and ∠
Let:
m and m
In the right ∆, we have:
⇒
⇒ m
Now, in the right ∆, we have:
⇒
⇒
On putting in the above equation, we get:
m
∴ Distance travelled by the ship during the period of observation = m
Page No 574:
Question 20:
The angle of elevation of the top of a building form a point A on the ground is 30°. On moving a distance of 30 m towards its base to a point B, the angle of elevation changes to 45°. Find the height of the building and the distance of its base form the point A.
Answer:
Let be the building and be the point on the ground making an angle of elevation on top of the building.
Thus, we have:
∠, ∠ and m
Let:
m and m
In the right ∆, we have:
⇒
⇒ m
Or,
m
In the right ∆, we have:
⇒
⇒
By putting in the above equation, we get:
⇒
⇒ m
∴ m
Distance of the base of the building from point = m
Height of the building = m
Page No 574:
Question 21:
The angle of elevation of the top of a tree from a point A on the ground is 60°. On walking 20 metres away from its base, to a point B, the angle of elevation changes to 30°.Find the height of the tree.
Answer:
Let be the tree and be the point on the ground such that ∠, m and ∠.
Let:
m and m
In the right ∆, we have:
⇒ (or)
In the right ∆, we have:
⇒
⇒
On putting , we get:
⇒
⇒
⇒ m
Hence, the height of the tree is 17.32 m.
Page No 574:
Question 22:
From the top of a-7-m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of the foot of the tower is 60°. Find the height of the tower.
Answer:
Let be the tower and be the building. Draw ⊥ .
We have:
m, ∠ and ∠
Let m and m such that m and m.
In the right ∆, we have:
⇒
⇒
Now, in the right ∆, we have:
⇒
⇒
By putting in the above equation, we get:
Hence, the height of the tower is 28 m.
Page No 575:
Question 23:
The angles of elevation of the top of a tower from two points at distances of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Show that the height of the tower is 6 metres.
Answer:
Let be the tower and be two points such that m and m.
Let:
m, ∠ and ∠
In the right ∆BCA, we have:
In the right ∆BDA, we have:
Multiplying equations (1) and (2), we get:
Height of a tower cannot be negative.
∴ Height of the tower = 6 m
Page No 575:
Question 24:
From a point P on the ground, the angle of elevation of a 10-m-tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P.
Answer:
Let AC be the building, P be the point on the ground and be the flagstaff.
Thus, we have:
m, ∠ and ∠
Let:
m and m
In the right ∆, we have:
⇒
⇒ = 17.32 m
In the right ∆, we have:
⇒
⇒
On putting in the above equation, we get:
⇒ m
Thus, we have:
Length of the flagstaff = m
Distance of the building from point P = m
Page No 575:
Question 25:
The angles of depression of the top and bottom of a 10 m high building from the top of a multi-storey building are 30° and 45° respectively. Find the height of the multi-storey building are 30° and 45° respectively. Find the height of the multi-storey building and the distance between the two buildings.
Answer:
Let be the multistorey building and be the high building such that m. Draw ⊥ .
Thus, we have:
∠ and ∠
Let m such that m and m such that m.
In the right ∆, we have:
⇒
⇒
Or,
Now, in the right ∆, we have:
⇒
⇒
By putting in the above equation, we get:
⇒
⇒ m
We have:
Height of the multistorey building = m
Distance between the two buildings = m
Page No 575:
Question 26:
From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 2.5 m from the banks, find width of the river.
Answer:
Let and be two points on the banks on the opposite side of the river and be the point on the bridge at a height of 2.5 m.
Thus, we have:
m, ∠PAD and ∠
In the right ∆, we have:
⇒
⇒ m
In the right ∆, we have:
⇒
⇒ m
∴ Width of the river = m
Page No 575:
Question 27:
Two men are on opposite sides of a tower. they measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 metres, find the distance between the two men.
Answer:
Let be the tower and be the positions of the two men standing on the opposite sides. Thus, we have:
∠, ∠ and m
Let m and m such that m.
In the right ∆, we have:
⇒
⇒ m
In the right ∆, we have:
⇒
⇒
On putting in the above equation, we get:
⇒ m
∴ Distance between the two men = m
Page No 575:
Question 28:
The horizontal distance between two towers is 60 metres. The angle of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 90 metres, find the height of the first tower.
Answer:
Let be the first tower and be the second tower.
Now, m and m such that m and ∠.
Let m such that m and m.
In the right ∆, we have:
⇒
⇒
⇒
⇒ = m
∴ Height of the first tower = m
Page No 575:
Question 29:
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of each pole and the distances of the point form the poles.
Answer:
Let and be the two poles of equal heights such that m.
∴ Distance between the two poles m
We have:
∠ and ∠
Let m such that m.
In the right ∆, we have:
⇒
⇒ ...(i)
Now, in the right ∆, we have:
⇒
⇒
⇒ ...(ii)
From (i) and (ii), we have:
⇒
⇒
⇒ m
∴ m
We now have:
Height of each pole = m
Distance of the pole from point O = m
Distance of the pole from point O = m
Page No 575:
Question 30:
As observed form the top of a 75-m-tall lighthouse, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Answer:
Let be the lighthouse such that m and and be the positions of the two ships.
Thus, we have:
∠ and ∠
Let m and m such that m.
In the right ∆, we have:
⇒
⇒ m
In the right ∆, we have:
⇒
= m
∴ Distance between the two ships = m
Page No 575:
Question 31:
A man on the deck of a ship, 16 m above water level observes that the angles of elevation and depression respectively of the top and bottom of a cliff are 60° and 30°. Calculate the distance of the cliff from the ship and height of the cliff.
Answer:
Let be the deck of the ship above the water level and be the cliff.
Now,
m such that m and ∠ and ∠.
If AD = x m and m, then m.
In the right ∆, we have:
⇒
⇒ m
In the right ∆, we have:
⇒
⇒
⇒ [∵ ]
⇒ m
∴ Distance of the cliff from the deck of the ship = m
And,
Height of the cliff = m
Page No 576:
Question 32:
A man on a cliff observes a boat at an angle of depression of 30° which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be 60°. Find the time taken by the boat to reach the shore.
Answer:
Let be the cliff and and be the two positions of the boat.
We have:
∠ and ∠
Let the speed of the boat be metres per minute.
Also,
Distance covered by the boat in six minutes
m
Suppose the boat takes minutes to reach from .
∴ m
Let:
m
In the right ∆, we have:
⇒
⇒ ...(i)
From the right ∆, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we have:
⇒
⇒
minutes
Hence, the required time is three minutes.
Page No 576:
Question 33:
A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of 30°. A girl standing on the roof of a 20 m high building, finds the angle of elevation of the same bird to be 45°. Boy and the girl are on the opposite sides of the bird. Find the distance of the bird from the girl.
Answer:
Let be the position of the bird, be the position of the boy and be the roof on which is the position of the girl.
Let:
⊥ and ⊥
Thus, we have:
m, ∠, m and ∠.
Let:
m
From the right ∆OLB, we have:
⇒
⇒ m
∴ m
From the right ∆, we have:
⇒
⇒ m
∴ Distance of the bird from the girl = m
Page No 576:
Question 34:
A 1.5-m-tall boy is standing at some distance from a 30-m-tall building. The angle of elevation from his eyes to the top of the building increase form 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Answer:
Let be the building and and be the two positions of the boy. Draw ∥ .
Thus, we have:
m, m, ∠ and ∠ and m
Now, in ∆, we have:
⇒
⇒ m
In ∆, we have:
⇒
⇒ m
∴ m = 32.87 m
Hence, the distance walked by the boy towards the building is m.
Page No 576:
Question 35:
The angle of elevation of the top of an unfinished tower at a distance of 75 m form its base is 30°. How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60°?
Answer:
Let be the unfinished tower, be the raised tower and O be the point of observation.
We have:
m, ∠ and ∠
Let m such that m.
In ∆AOB, we have:
⇒
⇒ m = m
In ∆, we have:
⇒
⇒ m
∴ Required height = m = 86.6 m
Page No 579:
Question 1:
At a certain instant the ratio of the lengths of a pillar and its shadow are in the ratio . At that instant, the angle of elevation of the sum is
(a) 30°
(b) 45°
(c) 60°
(d) none of these
Answer:
(a) 30°
Let be the pillar and be its shadow.
Let:
∠ and
In the right ∆, we have:
⇒ [∵ ]
Hence, the angle of elevation of the sun is .
Page No 579:
Question 2:
At a certain instant, the altitude of the sum is 60°. At that instant, the length of the shadow of a vertical tower is 100 m. The height of the tower is
(a)
(b)
(c)
(d)
Answer:
(b)
Let be the tower and be its shadow.
We have:
∠ and m
Let:
m
In the right ∆, we have:
⇒
⇒ m
Hence, the height of the tower is m.
Page No 579:
Question 3:
The height of a tower is . The angle of elevation of its top from a point 100 m away from its foot is
(a) 30°
(b) 45°
(c) 60°
(d) none of these
Answer:
(c) 60°
Let be the tower and be the point of observation.
We have:
m and m
In ∆, we have:
⇒
⇒
⇒
⇒
Hence, the angle of elevation is .
Page No 579:
Question 4:
The angle of elevation of the top of a tower from a point on the ground 30 m away from the foot of the tower is 30°. The height of the tower is
(a) 30 m
(b)
(c) 20 m
(d)
Answer:
(b)
Let be the tower and be the point of observation.
Also,
∠ and m
Let:
m
In ∆, we have:
⇒
⇒ m
Hence, the height of the tower is m.
Page No 579:
Question 5:
The string of a kite is 100 m long and it makes an angle of 60° with the horizontal. If these is no slack in the string, the height of the kite from the ground is
(a)
(b)
(c)
(d) 100 m
Answer:
(a)
Let be the string of the kite and be the horizontal line.
If ⊥ , then m and ∠.
Let:
m
In the right ∆, we have:
⇒
⇒ m
Hence, the height of the kite is m.
Page No 579:
Question 6:
An observer 1.5 m tall 28.5 away from a tower and the angle of elevation of the top of the tower form the eye of the observer is 45°. The height of the tower is
(a) 27 m
(b) 30 m
(c) 28.5 m
(d) none of these
Answer:
(b) 30 m
Let be the observer and be the tower.
Draw ⊥ , Let metres. Then,
m , m and ∠.
= m.
In right ∆, we have:
⇒
⇒
⇒ m
Hence the height of the tower is m.
Page No 579:
Question 7:
The shadow of a vertical tower on level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. The height of the tower is
(a)
(b)
(c) 9 m
(d) 13 m
Answer:
(a)
Let be the tower and be its shadows.
Thus, we have:
∠ and ∠
If m, then m.
Let:
m.
In ∆, we have:
⇒
⇒ ... (i)
Now, in ∆, we have:
⇒
⇒ ...(ii)
On putting the value of from (i) in (ii), we get:
⇒
⇒
On multiplying the numerator and denominator by , we get:
m
Hence, the height of the tower is m.
Page No 580:
Question 8:
From the top of a hill, the angles of depression of two consecutive km stones due east are found to be 30° and 45°. The height of the hill is
(a)
(b)
(c)
(d)
Answer:
(b)
Let be the hill making angles of depression at points and such that ∠, ∠ and km.
Let:
km and km
In ∆, we have:
⇒ ⇒ ...(i)
In ∆, we have:
⇒ ...(ii)
On putting the value of taken from (i) in (ii), we get:
⇒
⇒
⇒
On multiplying the numerator and denominator of the above equation by , we get:
km
Hence, the height of the hill is km.
Page No 580:
Question 9:
An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 45° and 60°. The width of the river is
(a)
(b)
(c)
(d)
Answer:
(a)
Let be the position of the aeroplane and be the horizontal line.
Let be the line on the ground such that ⊥ and m.
It observes an angle of depression such that ∠ and ∠.
Also,
∠ and ∠ (Alternate angles)
In ∆, we have:
⇒
⇒ m
Now, in ∆, we have:
⇒
⇒ m
∴ Width of the river = m
Page No 580:
Question 10:
If the angles of elevation of the top of a tower form tow points at distances a and b from the base and in the same straight line with it are complementary, then the height of the tower is
(a)
(b)
(c)
(d)
Answer:
(b)
Let be the tower and and be the points of observation on .
Let:
∠, ∠ and m
Thus, we have:
and
Now, in the right ∆ABC, we have:
⇒ ...(i)
In the right ∆ABD, we have:
⇒ ...(ii)
On multiplying (i) and (ii), we have:
⇒ [ ∵ ]
⇒
⇒ m
Hence, the height of the tower is m.
Page No 580:
Question 11:
On the level ground, the angle of elevation of a tower is 30°. On moving 20 m nearer, the angle of elevation is 60°. The height of the tower is
(a) 10 m
(b)
(c) 15 m
(d)
Answer:
(b)
Let be the tower and and be the points of observation such that ∠, ∠, m and m.
Now, in ∆, we have:
⇒
⇒
In ∆, we have:
⇒
∴
⇒
⇒ ⇒
∴ Height of the tower AB = m
Page No 580:
Question 12:
If the elevation of the sun changes form 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high, is
(a) 7.5 m
(b) 15 m
(c)
(d)
Answer:
(c)
Let be the pole and and be its shadows.
We have:
∠, ∠ and m
In ∆, we have:
⇒ ⇒ m
Now, in ∆, we have:
⇒ ⇒ m
∴ Difference between the lengths of the shadows = m
Page No 580:
Question 13:
On the same side of a tower 300 m high, the angles of depression of two objects are 45° and 60° respectively. The distance between the poles is
(a) 127 m
(b) 117 m
(c) 217 m
(d) 473 m
Answer:
(a) 127 m
Let be the tower and be the positions of the objects such that m, ∠ and ∠.
Now, In ∆, we have:
= 1
⇒
⇒ m
In ∆, we have:
⇒
⇒ m
∴ m
Hence, the distance between the two poles is 127 m.
Page No 580:
Question 14:
The heights of two poles are 80 m and 65 m. If the line joining their tops makes an angle of 45° with the horizontal, then the distance between the poles is
(a) 15 m
(b) 22.5 m
(c) 30 m
(d) 7.5 m
Answer:
(a) 15 m
Let and be the poles of heights of 80 m and 65 m, respectively.
Draw ⊥ .
Now,
∠
Also,
m
In ∆, we have:
⇒
⇒ m
∴ m
Hence, the distance between the poles is 15 m.
Page No 580:
Question 15:
From the foot of a tower, the angle of elevation of the top of a column is 60° and from the top of the tower, the angle of elevation is 30°. If the height of the tower is 25 m, the height of the column is
(a) 35 m
(b) 42.5 m
(c) 37.5 m
(d) 27.5 m
Answer:
(c) 37.5 m
Let be the tower and be the column.
Draw ⊥ .
Given:
m, ∠ and ∠
Thus, we have;
m.
Let m.
In the right ∆, we have:
⇒ ⇒
In the right ∆, we have:
⇒ ⇒
Now,
⇒
⇒
⇒
⇒ m
∴ Height of the column = m
Page No 581:
Question 16:
A straight tree breaks due to storm and the broken part bends so that the top of the trees touches the ground making an angle of 30° with the ground. the distance from the foot of the tree to the point, where the top touches the ground is 10 m. The height of the tree is
(a)
(b)
(c)
(d)
Answer:
(a)
Let be the tree broken at point such that takes the position so that ∠ and m.
Let:
m and m. So, the height of the tree is m.
Now, in ∆, we have:
⇒
⇒ m
Again,
⇒
⇒ m
∴ Height of the tree = m
Page No 581:
Question 17:
An observer standing 50 m away from a building notices that the angles of elevation of the top and bottom of a flagstaff on the building are 60° and 45° respectively. The height of the flagstaff is
(a)
(b)
(c)
(d)
Answer:
(c)
Let be the tower, be the flagstaff and O be the position of the observer. Thus, we have:
m, ∠ and ∠
Now, in ∆, we have:
⇒
⇒ m
In ∆, we have:
⇒
⇒ m
∴ Height of the flagstaff = m
Page No 581:
Question 18:
A boat is being rowed away from a cliff 150 m high. At the top of the cliff the angle of depression of the boat changed from 60° to 45° in 1 minute.
If the speed of the boat is
(a) 4.31 km/hr
(b) 3.81 km/hr
(c) 4.63 km/hr
(d) 3.91 km/hr
Answer:
(b) 3.81 km/h
Let be the cliff and be the two positions of the ships. Thus, we have:
m, ∠ and ∠
Now, in ∆, we have:
⇒
⇒ m
In ∆, we have:
⇒
⇒ m
Now,
m
∴ m
Speed of the boat = m/s km/h km/h
Page No 581:
Question 19:
In a rectangle, the angle between a diagonal and a side is 30° and the length of this diagonal is 8 cm. the area of the rectangle is
(a) 16 cm2
(b)
(c)
(d)
Answer:
(c)
Let be the rectangle in which ∠ and cm.
In ∆, we have:
⇒
⇒ m
Again,
⇒
⇒ m
∴ Area of the rectangle = cm2
Page No 581:
Question 20:
If the length of shadow a pole on a level ground is twice the length of the pole, the angle of elevation of the sun is
(a) 30°
(b) 45°
(c) 60°
(d) none of these
Answer:
(d) None of these
Let be the pole and be its shadow such that .
Let:
∠
In ∆, we have:
⇒
⇒
It is different from and .
Page No 581:
Question 21:
A pole 6 m high casts a shadow long on the ground. At that instant, the sun's elevation is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(c) 60°
Let be the pole and be its shadow.
Let ∠, m and m.
In ∆, we have:
⇒
⇒
⇒
Hence, the angle of elevation of the sun is 60°.
Page No 581:
Question 22:
Two men standing on opposite sides of a flagstaff measure the angles of the top of the flagstaff as 30° and 60°. If the height of the flagstaff is 18 m, the distance between the men is
(a) 24 m
(b)
(c)
(d) 31.2 m
Answer:
(b)
Let be the flagstaff and be the positions of the two men. Thus, we have:
m,∠ and ∠
In ∆, we have:
⇒
⇒ m
In ∆, we have:
⇒
⇒ m
∴ m
Hence, the distance between the two men is m.
Page No 581:
Question 23:
The angled of elevation of an aeroplane flying vertically above the ground as observed from two consecutive stones 1 km apart are 45° and 60°. The height of the aeroplane from the ground is
(a)
(b)
(c)
(d)
Answer:
(d)
Let be the position of the aeroplane and and be the two stones (1 km apart) such that km.
Let be the perpendicular meeting produced at . Thus, we have:
∠ and ∠
Let:
km and km
In ∆, we have:
⇒
⇒ (or) ...(i)
In ∆, we have:
⇒
⇒ (or) ...(ii)
Putting the value of from (i) in (ii), we get:
⇒
⇒
⇒ km
Hence, the height of the aeroplane from the ground is km.
Page No 582:
Question 24:
Match the following columns:
Column I | Column II |
(a) The length of a shadow of a tower is times the height of the tower. The angle of elevation of the sun is........ . |
(p) 40 m |
(b) The angle of depression of the top of a tower at a point 40 m from its base 45°. The height of the tower is ........ . |
(q) 60° |
(c) The angle of elevation of the top of a tower from a point 15 m away from its base is 30°. The height of the tower is .......... . |
(r) 30° |
(d) At a point 14 m away from the base of a high pillar, the angle of elevation of its top is ........ . |
(s) |
Answer:
(a) Let the length of the shadow be m. Then, according to the given condition, m.
Let be the angle of elevation. Then, m and m.
In ∆, we have:
⇒
⇒
∴
(b) Let be the tower. Then, m and ∠.
Let:
m
In ∆, we have:
⇒ ⇒ m
∴
(c) Let be the tower. Then, m and ∠.
Let:
m
In ∆, we have:
⇒
⇒ m
∴
(d) Let be the pillar and be the angle of elevation. Thus, we have:
m and m
In ∆, we have:
⇒
⇒
∴
Page No 582:
Question 25:
Match the following columns:
Column I | Column II |
(a) From the top of a 10 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. The height of the tower is ....... . |
(p) 12.7 m |
(b) A tower is 30 m high. Its shadow is x metres shorter when the sun's altitude is 45° than when it is 30°. Then, x = .......... . |
(q) 27.3 m |
(c) From the top of a cliff 30 m high, the angles of depression of the top and bottom of a tower are 30° and 45° respectively. The height of the tower is ......... . |
(r) 69.2 m |
(d) Two men on either side of a temple 30 m high observe the angles of elevation of the top of the temple to be 30° and 60° respectively. The distance between the two men is ........ . |
(s) 21.9 m |
Answer:
(a) Let be the building and be the tower. Draw ⊥ .
Now, we have:
m , ∠ and ∠
From the right ∆, we have:
= 1
⇒ ⇒ m
From the right ∆, we have:
⇒ [∵ BE = AC]
⇒ = m
Height of the tower = m
∴ (a) - (q)
(b) Let be the tower such that m. Thus, we have:
∠ and ∠
From the right ∆, we have:
⇒
⇒ m
From the right ∆, we have:
⇒
⇒ m
Now,
m
∴ (b) - (s)
(c) Let be the cliff such that m and be the tower. Thus, we have:
∠ and ∠
If m, then m.
From the right ∆, we have:
⇒
⇒ m
From the right ∆, we have:
⇒
⇒ m
We know:
m
∴
⇒
⇒
⇒ m
Hence, the height of the tower is 12.7 m.
∴ (c) - (p)
(d) Let be the temple and be the points of the observer. Thus, we have:
m , ∠ and ∠
From the right ∆, we have:
⇒
⇒ m
From the right ∆, we have:
⇒
⇒ = m
Distance between the two men = m
∴ (d) - (r)
Page No 583:
Question 26:
Assertion (A)
If two towers of height h1 and h2 subtend angles of 60° and 30° respectively at the midpoint of the line joining their feet, then h1 : h2 = 3 : 1.
Reason (R)
Figure
In the given figure, we have:
(a) Both Assertion (A) and Reason (R) are true and (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.
Answer:
There are two towers and of heights and , respectively. Let be the distance between the two towers and O be the midpoint of .
Thus, we have:
m
Now, in ∆, we have:
⇒ ⇒
Again, in ∆, we have:
⇒ ⇒
Now,
∴
Hence, both assertion (A) and reason (R) are true and (R) is the correct explanation of (A).
Page No 584:
Question 27:
Assertion (A)
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Then, the height of the tree is
Reason (R)
The angle of elevation of the top of a tower as observed from a point on the ground is α and on moving a metres towards the tower, the angle of elevation is β. Then, the height of the tower is
(a) Both Assertion (A) and Reason (R) are true and (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.
Answer:
Let be the tree on the opposite side of the bank of the river and be the position of the man standing on the bank of the river.
Thus, we have:
m, ∠ and ∠
If m and m, then m.
Then, in the right ∆ABC, we have:
⇒
⇒
In the right ∆ABD, we have:
⇒
⇒
⇒ [∵ ]
⇒ ⇒ ⇒ m
Hence, assertion (A) is true.
Assertion (A):
Let:
m, and
Given:
Putting the values of and in the above equation, we get:
m
Hence, reason (R) is true.
Both assertion (A) and reason (R) are true, and (R) is the correct reason of assertion (A).
Page No 591:
Question 1:
The angle of elevation of a jet plane form a point P on the ground is 60°. After a flight of 15 second, the angle of elevation changes to 30°. If the jet plane is flying at a constant of , then the jet plane is
(a) 360 km/hr
(b) 720 km/hr
(c) 540 km/hr
(d) 270 km/hr
Answer:
(b) 720 km/h
Let be the position of the jet plane and be the point on the ground such that ∠, ∠ and m.
Let m and m.
In the right ∆, we have:
⇒
⇒ m
Now, in the right ∆, we have:
⇒
⇒
Putting the value of in the above equation, we get:
⇒ m
∴ Distance between the two points = m
m/s = km/h
Page No 592:
Question 2:
A man standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. The width of the river is
(a) 20 m
(b) 22 m
(c) 24 m
(d) 25 m
Answer:
(a) 20 m
Let be the tree and D be the position of the man on the opposite side of the tree. Thus, we have:
∠, ∠ and m
If m and m, then m.
In ∆, we have:
⇒ ⇒ ...(i)
In ∆, we have:
⇒
⇒
⇒ ...(ii)
From (i) and (ii), we have:
⇒
⇒ m
From (i), we get:
m
∴ Width of the river = m
Page No 592:
Question 3:
As observed from the top of a 75 m tall lighthouse, the angle of depression of two ships is 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, then the distance between the ships is
(a)
(b)
(c)
(d)
Answer:
(b)
Let be the lighthouse. From O, the angle of depression is observed.
We have:
∠, ∠ and m
Let m and m such that m.
In ∆, we have:
⇒ m
In ∆, we have:
⇒ ⇒ m
∴ Distance between the two ships = m
Page No 592:
Question 4:
If the angles of elevation of a tower from two points as distances a and b, where a > b from its foot and in the same straight line from it are 30° and 60°, then the height of the tower is
(a)
(b)
(c)
(d)
Answer:
(d)
Let be the tower and be the two points. Thus, we have:
, ∠ and ∠
Let:
m
In right ∆, we have:
In right ∆, we have:
On multiplying (1) and (2), we get:
Hence, the height of the tower is .
Page No 592:
Question 5:
A ladder 20 m long touches a wall at the height of 10 m. Find the angle made by the ladder with the horizontal.
Answer:
Let be the ladder such that m. It touches the wall at a height m.
Let be the angle.
In ∆, we have:
⇒
Hence, the angle made by the ladder with the horizontal is 30°.
Page No 592:
Question 6:
The angle of elevation of the top of a pillar at a distance of 18 m from its foot is 30°. Find the height of the pillar.
Answer:
Let be the pillar, be the foot of the person such that m. Thus, we have:
∠
From right ∆, we have:
⇒
⇒ m
∴ Height of the tower = m
Page No 592:
Question 7:
Find the angle of elevation of the sum when the length of of the shadow of a pole is times the height of the pole.
Answer:
Let be the pole and be its shadow.
Let and such that (given) and be the angle of elevation.
From ∆, we have:
⇒
⇒
⇒
Hence, the angle of elevation is .
Page No 592:
Question 8:
A Minar A casts a shadow 150 m long at the same time when a Minar B casts a shadow 120 m long on the ground. If the height of the Minar B is 80 m, find the length of Minar A.
Answer:
Let be the Minar and Minar , respectively. Let and be the shadows of Minar and Minar .
Thus, we have:
m, m and m
Let and be the angle of elevation of both Minars.
From ∆, we have:
From ∆, we have:
⇒ [ ]
⇒ m
∴ Length of Minar = m
Page No 592:
Question 9:
If two towers of height h1 and h2 subtend angles of 60° and 30° respectively at the midpoint of the line joining their feet, then show that h1 : h2 = 3 : 1.
Answer:
Let and be the two towers subtending angles , respectively.
Now,
∠ and ∠
Let be the midpoint of the line segment joining the two towers.
Let:
, and
From ∆, we have:
⇒
⇒ ...(i)
From ∆, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we get:
⇒
∴
Page No 592:
Question 10:
Find the height of a tower whose shadow is 10 m long when the sun's altitude is 45°.
Answer:
Let be the tower such that m and be its shadow such that m and ∠.
From ∆, we have:
⇒
⇒ m
Hence, the height of the tower is 10 m.
Page No 592:
Question 11:
From the top of a building 60 m high, the angles of depression of the top and bottom of a vertical lamp post are observed to be 30° and 60° respectively. Find the height of the lamp post.
Answer:
Let be the building and be the vertical lamppost.
Here,
m, ∠ and ∠
Let m such that m. Thus, m and m.
From ∆, we have:
⇒ =
⇒ ...(i)
From ∆, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we have:
⇒
⇒
⇒
⇒ m.
∴ Height of the lamppost = m
Page No 593:
Question 12:
An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at that instant.
Answer:
Let be the aeroplane flying vertically below the aeroplane at a height of 3125 m, that is, m.
The angles of elevation of both the aeroplanes from the ground are such that ∠ and ∠.
Let:
m and m
From the right ∆, we have:
⇒
⇒
From the right ∆, we have:
⇒
⇒
Putting the value of , we get:
⇒ m
∴ Distance between the two aeroplanes = m
Page No 593:
Question 13:
A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point O on the ground is 60° and the angle of depression of the point O from the top of the tower is 45°. Show that the height of the tower is
Answer:
Let be the tower and be the pole such that m.
∠ and ∠
Let:
m and m
In right ∆ we have:
⇒
⇒ x = h
Or,
In right ∆, we have:
⇒
Putting the value of in the above equation, we have:
⇒
⇒
⇒ m
∴ Height of the tower = m
Page No 593:
Question 14:
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5. From a point on the plane, the angles of elevation of the bottom and top of the flagstaff are respectively 30° and 60°. Show that the height of the tower is 2.5 m.
Answer:
Let be the vertical tower and be the vertical flagstaff such that m.
∠ and ∠
Let:
m and m
In right ∆, we have:
⇒
⇒
In right ∆, we have:
⇒
Putting the value in above equation, we get:
⇒
⇒
⇒ m
∴ Height of the tower = m
Page No 593:
Question 15:
The angle of elevation of the top of a hill at the foot of the tower is 60° and the angle of elevation of the top of the tower form the foot of the hill is 30°. If the tower is 50 m high, find the height of the hill.
Answer:
Let be the hill and be the tower.
Now,
m, ∠ and ∠
Let:
m and m
From right ∆, we have :
⇒
⇒ m
From ∆, we have:
⇒
⇒ m [ ]
Height of the hill = m
Page No 593:
Question 16:
A tree breaks due to storm and the broken part bends so that the top of tree touches the ground, making an angle of 30° with the ground. If the distance between the foot of the tree to the point, where the top touches the ground is 8 m, show that the height of the tree is
Answer:
Let be the tree broken at point so that takes the position such that ∠ and m.
Let:
m and m
From right ∆, we have:
⇒
⇒
Again, we have:
⇒
⇒
Height of the tree = m
Page No 593:
Question 17:
The angle of elevation of the top of a tower as observed from a point on the ground is α and on moving metres towards the tower, the angle of elevation is β. Prove that the height of the tower is
Answer:
Let be the tower and be the point on the ground such that ∠.
On moving towards the tower at point , we have and ∠.
If and , then .
From ∆, we have:
⇒ ...(i)
From ∆, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we have:
⇒
⇒
⇒
Hence, the height of the tower is .
Page No 593:
Question 18:
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angle of elevation of the bottom and the top of the flagstaff are α and β respectively. Prove that the height of the tower is
Answer:
Let be the tower and be the vertical flagstaff such that . Thus, we have:
∠ and ∠
Let:
and
In ∆, we have:
⇒
⇒
Or,
...(i)
In ∆, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we get:
⇒
⇒
⇒
⇒
Hence, the height of the tower is .
Page No 593:
Question 19:
The angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection is β. Prove that the distance of the cloud from the point of observation is
Answer:
Let be the surface of the lake and be the point vertically above such that m.
Let be the position of the cloud and be its reflection on the lake.
Let the height of the cloud be metres. Thus, we have:
m
Draw ⊥.
Thus, we have:
∠ and ∠
m
m and m
From the right ∆, we have:
⇒
⇒ ...(i)
From the right ∆, we have:
⇒
⇒ ...(ii)
From (i) and (ii), we get:
⇒
⇒
⇒ ...(iii)
Using the value of in (i), we get:
...(iv)
Now, to find , we have:
⇒
Putting the value of from (iv), we get:
Distance of the cloud =
Page No 594:
Question 20:
A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 30°. Show that the speed of the boat is metres per minute.
Answer:
Let be the light house and and be the positions of the two ships. Thus, we have:
∠
∠
m
Let:
m and m
Thus, we have:
m
In ∆, we have:
In ∆, we have:
∴ m
Distance travelled in two minutes = = m
metres per minute
Hence proved.
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