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Class 10 RD Sharma Solutions – Chapter 12 Some Applications of Trigonometry – Exercise 12.1 | Set 2

• Last Updated : 13 Dec, 2022

Question 27. A T.V. tower stands vertically on a bank of a river of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60Â°. From a point 20 m away this point on the same bank, the angle of elevation of the top of the tower is 30Â°. Find the height of the tower and the width of the river.

Solution:

Let us considered AB be the T.V tower of height â€˜hâ€™ m on the bank of river and

D be the point on the opposite side of the river.

So, the angle of elevation at the top of the tower is 30Â°

So, AB = h and BC = x

And, it is given that CD = 20 m

So,

In Î”ACB

tan 60o = AB/BC

âˆš3x = h

x = h/âˆš3

Now, in Î”DBA,

tan 30o = AB/BD

1/âˆš3 = h / (20 + x)

âˆš3h = 20 + x

âˆš3h – h/âˆš3 = 20

2h/âˆš3 = 20

h = 10âˆš3 m

And,

x = h/âˆš3

x = 10âˆš3/ âˆš3

x = 10

Hence, the height of the tower is 10âˆš3 m and width of the river is 10 m.

Question 28. From the top of a 7 m high building, the angle of elevation of the top of a cable is 60Â° and the angle of depression of its foot is 45Â°. Determine the height of the tower.

Solution:

Given that, the height of the building(AB) = 7 m

Let us considered the height of the cable tower is CD,

So, the angle of elevation of the top of the cable tower from the top of the building = 60Â°,

Angle of depression of the bottom of the building from the top of the building= 45Â°,

From the figure we conclude that,

ED = AB = 7 m

And,

CD = CE + ED

So, In Î”ABD,

AB/ BD = tan 45Â°

AB = BD = 7

BD = 7

Now, in Î”ACE,

AE = BD = 7

And, tan 60Â° = CE/AE

âˆš3 = CE/ 7

CE = 7âˆš3 m

So, CD = CE + ED = (7âˆš3 + 7)= 7(âˆš3 + 1) m

Hence, the height of the cable tower is 7(âˆš3 + 1)m

Question 29. As observed from the top of a 75 m tall lighthouse, the angles of depression of two ships are 30Â° and 45Â°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution:

Given that, the height of the lighthouse(AB) = 75m,

The angle of depression of ship 1, Î± = 30Â°,

The angle of depression of bottom of the tall building, Î² = 45Â°,

From the figure,

Let us considered distance between ships(CD) = x m,

So, In Î”ABD,

tanÎ± = AB/BD

tan 30o = 75/(x + BC)

x + BC = 75âˆš3             ……(i)

So, In Î”ABC,

tanÎ² = AB/BC

tan45o = 75/BC

BC = 75         ……(ii)

Now on substituting eq(ii) in eq(i), we get

x + 75 = 75âˆš3

x = 75(âˆš3 – 1)

Hence, the distance between the two ships is 75(âˆš3 â€“ 1) m

Question 30. The angle of elevation of the top of the building from the foot of the tower is 30Â° and the angle of the top of the tower from the foot of the building is 60Â°. If the tower is 50 m high, find the height of the building.

Solution:

Let’ us considered AB be the building and CD be the tower.

So, according to the question, given that

The angle of elevation of the top of the building from the foot of the tower = 30Â°,

And, the angle of elevation of the top of the tower from the foot of the building = 60Â°,

Height of the tower = CD = 50 m,

So, In Î”CDB,

CD/ BD = tan 60Â°

50/ BD = âˆš3

BD = 50/âˆš3 â€¦. (i)

Next in Î”ABD,

AB/ BD = tan 30Â°

AB/ BD = 1/âˆš3

AB = BD/ âˆš3

AB = 50/âˆš3/ (âˆš3)              (From eq(i))

AB = 50/3

Hence, the height of the building is 50/3 m.

Question 31. From a point on a bridge across a river, the angle of depression of the banks on the opposite side of the river is 30Â° and 45Â° respectively. If the bridge is at the height of 30 m from the banks, find the width of the river.

Solution:

According to the question, given that

The height of the bridge from the bank = 30 m

Let us considered A and B represent the points on the bank on opposite sides of the river.

So, AB is the width of the river.

Now, P is a point on the bridge which is 30 m high from the banks.

From the figure,

AB = AD + DB

In Î”APD,

Given that, âˆ A = 30o

So, tan 30o = PD/ AD

1/âˆš3 = PD/ AD

AD = 30âˆš3 m

Now, in Î”PBD,

âˆ B = 45o

So, tan 45o = PD/ BD

1 = PD/ BD

BD = PD

BD = 30 m

As we know that, AB = AD + DB = 30âˆš3 + 30 = 30(âˆš3 + 1)

Hence, the width of the river is 30(1 + âˆš3) m

Question 32. 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 60Â° and 30Â° respectively. Find the height of the poles and the distances of the point from the poles.

Solution:

According to the question, given that

The distance between the poles(BD) = 80 m

The angles of elevation to the top of the points is 60o and 30o

Let us considered AB and CD be the poles and O is the point on the road.

From the figure,

In Î”ABO,

AB/ BO = tan 60Â°

AB/ BO = âˆš3

BO = AB/ âˆš3         …….(i)

So, in Î”CDO,

CD/ DO = tan 30o

CD/ (80 â€“ BO) = 1/ âˆš3

âˆš3CD = 80 â€“ BO

âˆš3AB = 80 â€“ (AB/âˆš3)              (Since AB = CD and using eq(i))

3AB = 80âˆš3 â€“ AB

4AB = 80âˆš3

AB = 20âˆš3

So, BO = 20âˆš3/(âˆš3) = 20 m

And, DO = BD â€“ BO = 80 â€“ 20 = 60

Hence, the height of the poles is 20âˆš3 m and

the distances of the points from the poles are 20m and 60 m.

Question 33. A man sitting at a height of 20 m on a tall tree on a small island in the middle of a river observes two poles directly opposite each other on the two banks of the river and in line with the foot of the tree. If the angles of depression of the feet of the poles from a point at which the man is sitting on the tree on either side of the river are 60Â° and 30Â° respectively. Find the width of the river.

Solution:

From the given figure,

Let’s width of river = PQ = (x + y) m

Height of tree (AB) = 20 m

So, in Î”ABP,

tan 60o = AB/ BP

âˆš3 = 20/ x

x = 20/ âˆš3 m

In Î”ABQ,

tan 30o = AB/ BQ

1/ âˆš3 = 20/ y

y = 20âˆš3

So, (x + y) = 20/ âˆš3 + 20âˆš3

[20 + 20(3)]/ âˆš3 = 80/âˆš3

Hence, the width of the river is 80/âˆš3 m.

Question 34. A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 7m. From a point on the plane, the angle of elevation of the bottom of the flagstaff is 30Â° and that of the top of the flagstaff is 45Â°. Find the height of the tower.

Solution:

According to the question,

The length of the flagstaff = 7 m

And the angles of elevation of the top and bottom of the flagstaff from point D are 45Â° and 30Â° respectively.

Let us considered the height of the tower (BC) = h m

And, DC = x m

So, in Î”BCD

tan 30Â° = BC/DC

1/ âˆš3 = h/ x

x = hâˆš3    ………(i)

And, in Î”ACD

tan 45Â°  = AC/ DC

1 = (7 + h)/ x

x = 7 + h

hâˆš3 = 7 + h                       (from eq(i))

h(âˆš3 â€“ 1) = 7

h = 7/(âˆš3 â€“ 1)

Now, on rationalizing the denominator we get

h = 7(âˆš3 + 1)/ 2 = 7(1.732 + 1)/2 = 9.562

Hence, the height of the tower is 9.56 m

Question 35. The length of the shadow of a tower standing on the level plane is found to 2x meters longer when the sunâ€™s altitude is 30Â° than when it was 30Â°. Prove that the height of the tower is x(âˆš3 + 1) meters.

Solution:

According to the question,

CD = 2x, âˆ D = 30Â°, and âˆ C = 45Â°

Let us considered the height of the tower (AB) = h m,

And the distance BC = y m,

So, in Î”ABC

tan 45Â° = AB/BC

1 = h/y

y = h

So, in Î”ABD

tan 30Â° = AB/BD

1/âˆš3 = h/ (2x + y)

2x + y = âˆš3h

2x + h = âˆš3h

2x = (âˆš3 â€“ 1)h

h = 2x/ (âˆš3 â€“ 1) x (âˆš3 + 1)/ (âˆš3 + 1)

h = 2x (âˆš3 + 1)/(3-1) = 2x (âˆš3 + 1)/2 = x (âˆš3 + 1)

Hence, the height of the tower is x (âˆš3 + 1) m

Hence Proved

Question 36. A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30Â° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 meters. Find the height of the tree.

Solution:

Let us considered AC be the height of the tree which is (x + h) m

Given that, the broken portion of the tree is making an angle of 30Â° with the ground.

From the given figure,

In Î”BCD,

tan 30o = BC/ DC

1/âˆš3 = h/ 10

h = 10/ âˆš3

Now, in Î”BCD

cos 30o = DC/BD

âˆš3/2 = 10/x

x = 20/âˆš3 m

So,

x + h = 20/âˆš3 + 10/âˆš3 = 30/âˆš3

10âˆš3 = 10(1.732) = 17.32

Hence, the height of the tree is 17.32 m

Question 37. A balloon is connected to a meteorological ground station by a cable of length 215 m inclined at 60Â° to the horizontal. Determine the height of the balloon from the ground. Assume that there is no slack in the cable.

Solution:

Let us considered the height of the balloon from the ground = h m

Given that, the length of the cable = 215 m,

and the inclination of the cable is 60o.

So,

In Î”ABC

sin 60o = AB/ AC

âˆš3/2 = h/215

h = 215âˆš3/2 = 185.9m

Hence, the height of the balloon from the ground is 186m (approx).

Question 38. Two men on either side of the cliff 80 m high observe the angles of elevation of the top of the cliff to be 30Â° and 60Â° respectively. Find the distance between the two men.

Solution:

Let us considered CL be the cliff and A and B are two men on either side of the cliff

And they are making angles of elevation with C as 30Â° and 60Â°

It is given that, the height of cliff CL = 80 m

Let us assume AL = x and BL = y

Now in Î”ACL,

tan 30o = 80/x

x = 80âˆš3

Similarly, in Î”BCL

tan 60o = CL/BL = âˆš3 = 80/y

y = 80/âˆš3

Hence, the distance between two men,

x + y = 80âˆš3 + 80/âˆš3

240+80 / âˆš3 = 320/âˆš3

On rationalize the value, we get,

320âˆš3 / 3 = 184.746m

Question 39. Find the angle of elevation of the sun (sunâ€™s altitude) when the length of the shadow of a vertical pole is equal to its height.

Solution:

Let us considered the height of pole AB = h m,

Then the height of shadow = h m

Lets us considered the angle of elevation be Î¸,

So, In Î”ABC,

tan Î¸ = AB/BC = h/h = 1

tan Î¸ = 1 = tan 45o

Hence, Î¸ = 45o

Question 40. An airplane is flying at a height of 210 m. Flying at this height at some instant the angles of depression of two points in a line in opposite directions on both the banks of the river are 45Â° and 60Â°. Find the width of the river. (Use âˆš3  = 1.73)

Solution:

According to the question,

The height of the airplane = 210 m

Let us considered that, AB is the height of the airplane and

C and D are the points on the opposite banks of a river.

So, the angle of elevation of A are 45Â° and 60Â°

Let us considered CB = x and BD = y

In Î”ABD,

tan Î¸ = P/B

tan 45o = AB/BD = 1 = AB/BD

210/BD = 1

BD = 210 m

Similarly, In Î”ABC,

tan 60o = AB/CB = âˆš3 = 210/CB

CB = 210/âˆš3

On rationalize the value we get,

CB = 70âˆš3 m = 121.10m

So, the width of river = CB + BD

= 121.10 + 210

= 331.10m

Question 41. The angle of elevation of the top of a chimney from the top of a tower is 60Â° and the angle of depression of the foot of the chimney from the top of the tower is 30Â°. If the height of the tower is 40 m, find the height of the chimney. According to pollution control norms, the minimum height of a smoke-emitting chimney should be 100 m. State if the height of the above-mentioned chimney meets the pollution norms. What value is discussed in this question?

Solution:

Let us considered, CD be the tower and AB be the chimney

So, the angle of elevation of the top of the tower with the top of the chimney is 60Â° and

the foot of chimney with the top of the tower is 30Â°

So, it is given that CD = 40 m

From the given figure,

CE|| DB

So, CE = DB and EB = CD = 40

Now in â–³BCD

tan 30Â° = 40/DB = 1/âˆš3

DB = 40âˆš3             …….(i)

Now, in â–³ACE

tan 60Â° = AE/CE = âˆš3 = (h – 40)/DB

DB = (h-40)/âˆš3           …….(ii)

From eq(i) and (ii), we get

40âˆš3 = (h – 40)/âˆš3 = 120 + 40 = 160m

Hence, the height of the chimney is 160m.

Question 42. Two ships are there in the sea on either side of a lighthouse in such a way that the ships and the lighthouse are in the same straight line. The angle of depression of the two ships are observed from the top of the lighthouse are 60Â° and 45Â° respectively. If the height of the lighthouse is 200 m, find the distance between the two ships. (Use âˆš3 = 1.73)

Solution:

Let us considered that, AB be the lighthouse and

C and D are two ships which make an angle of depression

with the top A of the lighthouse of 60Â° and 45Â°

So, âˆ C = 45Â° and âˆ D = 60Â°

It is given that, AB = 20 m

Now in â–³ACB,

tan 45Â° = 20/CB

CB = 20m

Similarly, in â–³ABD,

tan 60Â° = AB/BD

âˆš3 = 20/BD

BD = 20/âˆš3 = 20âˆš3/3 m

Now, the Distance between two ships

CD = CB + BD

20 + 20/3 âˆš3 = 20(1 + âˆš3)/3 m

= 20 x 0.19 = 18.20m = 18.2m

Question 43. The horizontal distance between the two poles is 15 m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30Â°. If the height of the second pole is 24 m, find the height of the first pole. (âˆš3 = 1.732)

Solution:

Let us considered AB and CD are two poles and

the distance between them(BD) = 15 m,

The height of pole(AB) = 24 m

Angle of elevation from the top of pole CD, to pole AB = 30Â°

Now, from point C, draw CE || DB

Let us considered CD = x m,

CE = DB = 15m and AE = AB – EB

AE = AB – CD = (24 – x)m

Now in â–³ACE,

tan 30Â° = (24 – x)/15

1/âˆš3 = (24 – x)/15

15 = âˆš3 (24 – x)

x = 24 – 15/âˆš3

Now, after rationalizing we get,

x = 15.34m

Question 44. The angles of depression of two ships from the top of a lighthouse and on the same side are 45Â° and 30Â° respectively. If the ships are 200 m apart, find the height of the lighthouse.

Solution:

Let us considered PQ be the lighthouse and

A and B are two ships on the same side of the lighthouse

The angle of depression from the top of the lighthouse of the two ships are 30Â° and 45Â°

So, âˆ A = 30 and âˆ b = 45 and AB = 200m

Now, let us considered the height of lighthouse = h,

and BQ = x

Now in â–³PBQ

tan 45Â° = h/x

h = x

Similarly, in â–³PAQ,

tan 30Â° = h/(200 + x)

1/âˆš3 = h/(200 + h)

h(âˆš3 – 1) = 200

h = 200/0.732 m = 273.2m

Hence, the height of the house is 273.2m.

Question 45. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Solution:

Let us considered TR be the tower.

A and B are two points which make the angled elevation with top of the tower as Î¸ and 90Â° â€“ Î¸ (âˆµ angles are complementary)

Also, the height of tower TR = h and AR = 9 m, BR = 4m

Now in â–³TAR

tan Î¸ = TR/AR = h/9      ……(i)

Similarly, in right â–³TBR,

tan(90 – Î¸) = h/4 = cot Î¸ = h/4    ……(ii)

Now, on multiplying eq(i) and (ii), we get

tan Î¸ x cot Î¸ = h/4 x h/9

1 = h2/36

h2 = 36

h = 6m

Hence, height of tower is 6m

Question 46. From the top of a 50 m high tower, the angles of depression of the top and bottom of a pole are observed to be 45Â° and 60Â° respectively. Find the height of the pole.

Solution:

Let us considered AB be the tower and CD is the pole.

So, the angles of depression from the top A to the top and bottom of the pole are 45Â° and 60Â°

AB = 50 m,

Let us considered CD = h and BD = EC = x

âˆµ CE || DB

âˆ´ EB = CD = h

and AE = 50 â€“ h

tan 60Â° = 50/x

âˆš3 = 50/x

x = 50/âˆš3            ……(i)

Similarly, in â–³ACE,

tan 45Â° = AE/CE = 1 = (50 – h)/x

x = 50 – h

x + h = 50

h = 50 – x

50 – (50/âˆš3) = 21.13m             (using eq(i))

Hence, the height of pole is 21.13m

Question 47. The horizontal distance between two trees of different heights is 60 m. The angle of depression of the top of the first tree, when seen from the top of the second tree, is 45Â°. If the height of the second tree is 80 m, find the height of the first tree.

Solution:

Let us considered AB and CD be the two trees

So, given that AB = 80 m, angle of depression from A to C is 45Â°.

Now, draw CE || DB

âˆ ACE = âˆ XAC = 45Â°

Let us assume CD = h

CE = BD = 60 m

EB = CD = h and AE = 80 – h

Now in right â–³ACE

tan 45Â° = (80 – h)/60

80 – h = 60

h = 20m

Hence the height of the second tree is 20m.

Question 48. A flag-staff stands on the top of a 5 m high tower. From a point on the ground, the angle of elevation of the top of the flag-staff is 60Â° and from the same point, the angle of elevation of the top of the tower is 45Â°. Find the height of the flagÂ­staff.

Solution:

Let us considered FT is the flag-staff situated on the top of the tower TR.

Now, A is any point on the same plane that makes elevation angles with

top of the flag-staff and top of the tower are 60Â° and 45Â°.

Given that the length of tower TR = 5m

Let us assume that the height of flag-staff FT = h and AR = x, then

In â–³TAR,

tan 45Â° = 5/x

x = 5           …..(i)

Similarly in â–³FAR,

tan 60Â° = FR/AR = âˆš3 = (h + 5)/x

x = (h + 5)/âˆš3           ……(ii)

Now, from eq(i) and (ii), we get

5 = (h + 5)/âˆš3

5âˆš3 = h + 5

h = 5(âˆš3 – 1) = 3.65m

Hence the height of the flag-staff is 3.65m.

Question 49. The angle of elevation of the top of a vertical tower PQ from a point X on the ground is 60Â°. At a point Y, 40 m vertically above X, the angle of elevation of the top is 45Â°. Calculate the height of the tower.

Solution:

Let us considered that TR is the tower .

Now, from a point X, the angle of elevation of T is 60Â° and 40 m

above X, from the point Y, the angle of elevation is 45Â°

From Y, draw YZ || XR

Let us considered, TR = h and XR = YZ = x

ZR = YX = 40m and TZ = (h – 40) m

In â–³TXR,

tan 60Â° = h/x

âˆš3 = h/x

x = h/âˆš3              ……(i)

Similarly in â–³TYZ,

tan 45Â° = TZ/YZ

1 = (h – 40)/x

x = h – 40              ……(ii)

From eq(i) and (ii), we get

h/âˆš3 = h – 40

h(âˆš3 – 1) = 40âˆš3

h = 40âˆš3/(âˆš3 – 1)

On rationalize above term and we get,

h = 94.64m

Hence, the height of the tower is 94.64m

Question 50. As observed from the top of a 150 m tall lighthouse, the angles of depression of two ships approaching it are 30Â° and 45Â°. If one ship is directly behind the other, find the distance between the two ships.

Solution:

Let us considered, LH be the lighthouse,

So, A and B are the two ships making angles of elevation

with the top of the lighthouse as 30Â° and 45Â°.

LH = 150 m.

Let us considered AB = x and BH = y

Now in â–³LAH,

tan 30Â° = 150/(x + y)

1/âˆš3 = 150/(x + y)            ……..(i)

Similarly in â–³LBH,

tan 45Â° = LH/BH

1 = 150/y

y = 150  ……(ii)

From eq(i), we get

x + 150 = 150âˆš3

x = 150âˆš3 – 150

= 109.5 m

Hence the Distance between two ships is 109.5 m.

Question 51. The angles of elevation of the top of the rock from the top and foot of a 100 m high tower are respectively 30Â° and 45Â°. Find the height of the rock.

Solution:

Let us considered RS is the rock and TP is the tower.

So, the angles of elevation of the top of the rock with

the top and foot of the tower are 30Â° and 45Â°.

Given that, the height of TP = 100 m

Let the height of rock RS = h

From T, draw TQ || PS

Then QS = TP = 100 m

and RQ = h â€“ 100

Let us assume, PS = TQ = x

Now in right Î”RPS,

tan 45Â° = h/x

x = h  …..(i)

Similarly in â–³RTQ,

tan 30Â° = RQ/TQ = 1/âˆš3 = (h – 100)/x

x = âˆš3(h – 100)     ….(ii)

From eq(i) and (ii), we get

h = âˆš3(h – 100)

h(âˆš3 – 1) = 100âˆš3

On rationalize above value we get,

h = 50âˆš3(âˆš3 – 1) = 236.5 m

Hence the height of rock is 236.5 m.

Question 52. A straight highway leads to the foot of a tower of height 50 m. From the top of the tower, the angles of depression of two cars standing on the highway are 30Â° and 60Â° respectively. What is the distance between the two cars and how far is each car from the tower?

Solution:

Let us considered, TR be the tower

And A and B are two cars on the road making angles of elevation

with T the top of the tower as 30Â° and 60Â°

Height of the tower TR = 50 m

Let us considered AR = x and BR = y

Now in Î”TAR,

tan 30Â° = TR/AR = 50/x

1/âˆš3 = 50/x

x = 50âˆš3         …….(i)

Similarly in â–³TRB,

tan 60Â° = TR/BR = 50/y

âˆš3  =50/y

y = 50/âˆš3       …..(ii)

(i) Now we find the distance between the two cars

AB = AR – BR = x – y

= (50âˆš3 – 50/âˆš3) = 50(âˆš3 – 1/âˆš3)

= 50((3 – 2) / âˆš3) x (50 x 2)/âˆš3

On rationalize above term and we get,

AB = 57.7 m

(ii) Now we find the how far is each car from the tower

Distance of A car = x = 50âˆš3 = 86.65 m

Distance of B car = y = 50/âˆš3 = 50âˆš3/3 = 28.83 m

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