# Frank Solutions for Class 10 Maths Chapter 22 Heights and Distances

Frank Solutions for Class 10 Maths Chapter 22 Heights and Distances are the best study materials for those students who have difficulties in solving problems. Students can download the PDF of Frank Solutions for Class 10 Maths Chapter 22 from the accessible links. Frank Solutions include all the questions provided in the textbook that is prescribed for Class 10, in accordance with the ICSE Board.

Chapter 22 – Heights and Distances explains that height is the measurement of an object in the vertical direction, and distance is the measurement of an object from a particular point in the horizontal direction. This chapter deals with some special techniques which help in solving problems easily and quickly. These are basic identities which we use in higher studies.

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1. The distance of the gate of a temple from its base is âˆš3 times it height. Find the angle of elevation of the top of the temple.

Solution:-

Let us assume that PQ be the temple and R be the position of its gate.

Let us assume h be the height of the temple,

Then,

PQ = h

QR = Distance of the gate of temple from its base = âˆš3h

In Î”PQR,

tan Î¸ = PQ/QR

tan Î¸ = h/âˆš3h

tan Î¸ = 1/âˆš3

So,

tan 30o = 1/âˆš3

Therefore, Î¸ = 30o

Hence, the angle of elevation of the top of the temple is 30o.

2. The angle of elevation of the top of a vertical cliff from a point 30 m away from the foot of the cliff is 60o. Find the height of the cliff.

Solution:-

Let us assume that PQ be the cliff and angle of elevation from R is 60o.

In Î”PQR,

tan Î¸ = PQ/QR

tan 60o = PQ/30

âˆš3 = PQ/30

PQ = 30âˆš3 m

Therefore, height of cliff is 30âˆš3 m.

3. A vertical pole is 90 m high and the length of its shadow is 90âˆš3 m. What is the angle of elevation of the sun?

Solution:-

Let us assume PQ be the pole and QR be its shadow.

Then,

In Î”PQR,

tan Î¸ = PQ/QR

tan Î¸ = 90/90âˆš3

tan Î¸ = 1/âˆš3

So,

tan 30o = 1/âˆš3

Therefore, Î¸ = 30o

Hence the angle of elevation is 30o.

4. The length of the shadow of a pillar is 1/âˆš3 times the height of the pillar. Find the angle of elevation of the sun.

Solution:-

Let us assume PQ be the pillar and QR be its shadow.

Let us assume h be the height of the pillar.

Then,

QR = (1/âˆš3)h

In Î”PQR,

tan Î¸ = PQ/QR

tan Î¸ = h/(h/âˆš3)

tan Î¸ = âˆš3

So,

tan 60o = âˆš3

Therefore, Î¸ = 60o

Hence the angle of elevation is 60o.

5. Find the length of the shadow cast by a tree 60 m high when the sunâ€™s altitude is 30o.
Solution:-

Let us assume that PQ be the tree of height 60 m and QR be its shadow.

Then,

In Î”PQR,

tan 30o = PQ/QR

(1/âˆš3) = 60/QR

QR = 60âˆš3 m

Therefore, height of tower is 60âˆš3 m.

6. An aeroplane takes off at angle of 30o with the ground. Find the height of the aeroplane above the ground when it has travelled 386 m without changing direction.

Solution:-

As per parameters given in the question,

Let us assume the plane takes off from point R on the ground and also assume that P be the final position of the plane.

Then,

In Î”PQR,

sin 30o = PQ/PR

Â½ = PQ/386

PQ = 386/2 m

PQ = 193 m

Therefore, the required height of the aeroplane above the ground is 193 m.

7. A kite tied to a string makes an angle of 60o with the ground. Find the perpendicular height of the kite if the length of its string is 250 m.
Solution:-

Let us assume that P is the kite and the string is tied to point R on ground.

Then,

In Î”PQR,

sin 60o = PQ/PR

âˆš3/2 = h/250

h = 250âˆš3/2 m

h = 125âˆš3 m

Therefore, the required perpendicular height of the kite is 125âˆš3 m.

8. The topmost branch of a tree is tied with a string attached to a pole in the ground. The length of this string is 200 m and it makes an angle of 45o with the ground. Find the distance of the pole to which the string is tied from the base of the tree.

Solution:-

As per parameters given in the question,

Let us assume that the topmost branch of the tree is at point Q and R is the point on the ground to which the topmost branch is tied.

In Î”PQR,

cos 45o = PR/PQ

1/âˆš2 = h/200

h = 200/âˆš2 m

h = 200âˆš2/2 m

h = 100âˆš2

h = 100 Ã— 1.414

h = 141.4

Therefore, the required distance is 141.4 m.

9. The top of a ladder reaches a point on the wall 5 m above the ground. If the foot of the ladder makes an angle of 30o with the ground, find the length of the ladder.

Solution:-

As per parameters given in the question,

Let us assume that PR be the ladder and PQ is the point which is 5 m above the ground.

Then,

In Î”PQR,

sin 30o = PQ/PR

Â½ = 5/PR

PR = 10 m

Therefore, the length of the ladder is 10 m.

10. A 10 m high pole is kept vertically by a steel wire. The wire is inclined at an angle of 40o with the horizontal ground. If the wire runs from the top of the pole to the point on the ground where its other end is fixed, find the length of the wire.

Solution:-

As per parameters given in the question,

Let us assume that PQ be the pole and PR be the wire which runs from the top of the pole to the point on the ground where its other end is fixed.

Then,

In Î”PQR,

sin 40o = PQ/PR

0.6428 = 10/PR

PR = 10/0.6428

PR = 15.6 m

Therefore, the length of the wire is 15.6 m.

11. A ladder rests against a tree on one side of a street. The foot of the ladder makes an angle of 50o with the ground. When the ladder is turned over to rest against another tree on the other side of the street it makes an angle of 40o with the ground. If the length of the ladder is 60 m, find the width of the street.

Solution:-

As per parameters given in the question,

Let us assume that PQ and RS be two trees and O be a point on the street AC between the two trees.

Then,

OS and OQ denotes the ladder at the two instants.

Then,

In Î”ORS,

cos 50o = OR/OS

0.6428 = OR/60

OR = 0.6428 Ã— 60 = 38.568

Now, consider Î”PQO,

cos 40o = PO/QO

0.7660 = PO/60

PO = 0.7660 Ã— 60 = 45.96

Therefore, PR = PO + OR

PR = 45.96 + 38.568

PR = 84.528 m

Hence the width of the street is 84.528 m.

12. Due to a heavy storm, a part of a banyan tree broke without separating from the main. The top of the tree touched the ground 15 m from the base making an angle of 45o with the ground. Calculate the height of the tree before it was broken.

Solution:-

As per parameters given in the question,

Let us assume that PR was original tree. Due to storm it was broken into two parts. The broken part Pâ€™ Q is making 45o with ground.

In Î”Pâ€™QR,

tan 45o = QR/Pâ€™R

1 = QR/15

QR = 15

Then,

Pâ€™R/Pâ€™Q = cos 45o

15/Pâ€™Q = 1/âˆš2

Pâ€™Q = 15âˆš2

Therefore, height of tree = QR + Pâ€™Q

= 15 + 15âˆš2

= 15 (1 + âˆš2)

= 15 Ã— 2. 414

= 36.21 m

Hence, the height of tree was 36.21 m.

13. The top of a palm tree having been broken by the wind struck the ground at an angle of 60o at a distance of 9 m from the foot of the tree. Find the original height of the palm tree.

Solution:-

As per parameters given in the question,

Let us assume that PR was original tree. It was broken into two parts. The broken part Pâ€™Q is making 60o with ground.

In Î”Pâ€™QR,

tan 60o = QR/Pâ€™R

âˆš3 = QR/9

QR = 9âˆš3

Then,

Pâ€™R/Pâ€™Q = cos 60o

9/Pâ€™Q = Â½

Pâ€™Q = 18

So, height of tree = QR + Pâ€™Q

= 9âˆš3 + 18

= (9 Ã— 1.732) + 18

= 15.588 + 18

= 33.588

Therefore, the height of tree was 33.588 m.

14. Find the angle of depression from the top of a 140 m high pillar of a milestone on the ground at a distance of 200 m from the foot of the pillar.

Solution:-

As per parameters given in the question,

Let us assume that PQ be the pillar and also angle of depression be Î¸.

Consider the Î”PQR,

tan Î¸ = PQ/RQ

tan Î¸ = 140/200

tan Î¸ = 7/10

tan Î¸ = 0.7

We have, tan 35o = 0.7

Therefore, the angle of depression is Î¸ = 35o.

15. The height of an observation tower is 180 m above sea level. A ship coming towards the tower is observed at an angle of depression of 30o. Calculate the distance of the boat from the foot of the observation tower.

Solution:-

As per parameters given in the question,

Let us assume that PQ be the tower and R be the position of the ship.

Also assume the distance of the boat from the foot of the observation tower be x.

Consider the Î”PQR,

tan Î¸ = PQ/RQ

tan 30o = 180/x

1/âˆš3 = 180/x

x = 180âˆš3

x = 180 Ã— 1.732

x = 311.76

Therefore, the distance of the boat from the foot of the observation tower is 311.76 m.