Exercise 9.1

1. A circus artist is climbing a 20 m long rope, which istightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, ifthe angle made by the rope with the ground level is 30° (see Fig).

Solution:

Given:

Length of rope = m

The angle of rope with ground = °= ∠ACBBAD

We have to find the height of the pole.

AB = Height of the Pole

BC = Distance between the point on the ground and the pole.

AC = Length of the Rope (Hypotenuse)

In ΔABC,

sinC = ABACBCAC

sin30° = AB20AC20

12 = AB20 [Since sin30° = 12]

AB = (12) ×

AB = m

Height of pole AB = 10 m

2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m Find the height of the tree.

Solution:

Height of the tree = AB +

AB, BC and ∠C is tan θ, where AB can be measured.

AB, AC and ∠C is sin θ, where AC can be measured.

Distance between the foot of the tree to the point where the top touches the ground = BC = m

In triangle ABC,

tan C = ABBCABAC

tan 30° = AB8

13 = AB8 [Since tan 30° = 13]

AB =83

sin C = ABACBCAC

sin 30° = 83AC

12 = 83 × 1AC

AC = 83 × 212

AC = 163

Height of tree = AB + AC

= 83 + 163

= 243

= 24×33×3 [On rationalizing ]

= 2433

= 8343

So, the height of tree is 83 meters.

3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution:

4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Solution:

We have to find the height of the tower.

Let us consider the height of the tower as AB, the distance between the foot of the tower to the point on the ground as BC.

In ΔABC, trigonometric ratio involving AB, BC and ∠C is tan θcos θ.

tan C = ABBCABAC

tan 30° = AB30BC30

133 = AB30

AB = 3033013

= 30×33×3

= 3033

= 1033

Height of tower AB = 103 m

5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Solution:

We take the height of the flying kite as AB, the length of the string as AC, and the inclination of the string with the ground at ∠C.

Trigonometric ratio involving AB, AC and ∠C is sinθcosθ.

In ΔABC,

sinC = ABACBCAC

sin 60° = 60AC60AB

32−32 = 60AC

AC = ( × 23)

= 120×33×3

= 12033

= 403203

Length of the string AC = 403m.

6. 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 increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Solution:

7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively.Find the height of the tower.

Solution:

Let the height of the building is BC while the height of the transmission tower which is fixed at the top of the building be AB.

D is the point on the ground from where the angles of elevation of the bottom B and the top A of the transmission tower AB are 4560° and 6090° respectively.

The distance of the point of observation D from the base of the building C is CDBC.

Combined height of the building and tower = AC = AB + BCAD

Trigonometric ratio involving sides AC, BC, CD, and ∠D (45° and 60°) is tan θcos θ.

In ΔBCD,

tan 45° = BCCDCDBC

= 20CD

CD =

In ΔACD,

tan 60° = ACCDCDAC

313 = AC20

AC = 3

Height of the tower, AB = AC - BC

AB = 203 - m

= 20 (3 - ) m

8. A statue, 1.6 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.

Solution:

Let the height of the pedestal be BC, the height of the statue, which stands on the top of the pedestal, be AB. D is the point on the ground from where the angles of elevation of the bottom B and the top A of the statue AB are 45° and 60° respectively.

The distance of the point of observation D from the base of the pedestal is CD. Combined height of the pedestal and statue AC = AB + BC

Trigonometric ratio involving sides AC, BC, CD, and ∠D (45° and 60°) is tan θcos θ.

In ΔBCD,

tan 45° = BCCDCDBC

= BCCD

Thus, BC =

In ΔACD,

tan 60° = ACCDCDAC

tan 60° = AB+BCCD

3 = 1.6+BCBC [Since BC = CD]

3 BC = + BC

3 BC - = 1.6

BC (3 - 1) =

BC = 1.6×3+13−13+1

= 1.63+13−1

= 1.63+12

= 0.83+10.63+1

Height of pedestal BC =0.83+1 m.

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

Solution:

Let the height of the tower be AB and the height of the building be CD.

The angle of elevation of the top of building D from the foot of tower B is 3060° and the angle of elevation of the top of tower A from the foot of building C is 6030°.

Distance between the foot of the tower and the building is BC.

Trigonometric ratio involving sides AB, CD, BC and angles ∠B and ∠C is tan θ.

In ΔABC,

tan 60° = ABBCBCAB

3 = 50BC30BC

BC = 503 ....(i)

In ΔBCD,

tan 30° = CDBCBCCD

13−13 = CDBC

13 = CD503 [from (i) BC = 503]

CD = 13 × 503

CD = 503503

Height of the building CD = 503 m.

10. 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 the poles and the distances of the point from the poles.

Solution:

Let us consider the two poles of equal heights as AB and DC and the distance between the poles as BC.

From a point O, between the poles on the road, the angle of elevation of the top of the poles AB and CD are 60° and 30° respectively.

Trigonometric ratio involving angles, distance between poles and heights of poles is tan θ.

Let the height of the poles be x

Therefore AB = DC = x

In ΔAOB,

tan 60° = ABBOBOAB

3 = xBO

BO = x3 ....(i)

In ΔOCD,

tan 30° = DCOCOCDC

13 = xBC−OB

13 = x80−x3 [from (i)]

- x3 = 3x

x3 + 3x =

x (13 + 313) = 80

x1+33 =

x 4334 = 80

x = 8034

x = 3

Height of the poles x = 203 m.

Distance of the point O from the pole AB

BO = x3

= 2033

=

Distance of the point O from the pole CD

OC = BC - BOOC

= 80 -

=

The height of the poles is 20√3 m and the distance of the point from the poles is 20 m and 60 m.

11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig). Find the height of the tower and the width of the canal.

Solution:

Considering ΔABC,

tan 60° = ABBCBCAB

313 = ABBC

AB = BC313 ...(i)

Considering ΔABD,

tan 30° = ABBDBDAB

tan 30° = ABCD+BC

13 = BC320+BC [from (i)]

+ BC = BC3 × 3

20 + BC = × BC

3BC - BC =

BC = 20

BC =

Substituting BC = 10 m in Equation (1), we get AB = 3 m

Height of the tower AB = 103 m

Width of the canal BC = m

12. 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 its foot is 45°. Determine the height of the tower.

Solution:

Let the height of the tower be CE and the height of the building be AB. The angle of elevation from the top E of the tower to the top A of the building is 60° and the angle of depression from the bottom C of the tower to the top A of the building is 45°.

Draw AD || BC.

Then, ∠DAC = ∠ACB = 45° (alternate interior angles)

In ΔABC,

tan 45° = ABBCBCAB

= 7BC

BC =

ABCD is a rectanglesquare.

Therefore, BC = AD = and AB = CD =

In ΔADE,

tan 60° = EDADADED

313 = ED7AD7

ED = 7337

Height of tower = CE = ED + CD

= 73 +

= 7 (3 + )

Height of the tower = 7 (3 + 1) m.

13. As observed from the top of a 75 m high lighthouse from the sea-level, 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.

Solution:

Let the height of the lighthouse from the sea level be AB and the ships are C and D.

The angles of depression of the ships C and D from the top A of the lighthouse are 30° and ° respectively.

Distance between the ships = CD = BD − BCAC

In ΔABC,

tan 45° = ABBCBCAB

= 75BC

BC =

In ΔABD,

tan 30° = ABBDBDAB

13 = 75BD−75BD

BD = 753

Distance between two ships CD = BD - BC

CD = 753 -

= 75 (3 - )

Distance between two ships CD is 75 (3 - 1) m.

14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13).Find the distance travelled by the balloon during the interval.

Solution:

Trigonometric ratio involving , BC, OD, and angles is tan θcos θ.

Distance travelled by the balloon OB = AB - OAOD

From the figure, OD = BCAB, and it can be calculated as

88.2 m - 1.2 m = m --- (1)

In ΔAOD,

tan 60° = ODOAOAOD

3 = 87OA87OD

OA = 8738713

= 87×33×3

= 87×33

= 3 m

In ΔABC,

tan 30° = BCABABBC

13 = 87AB87AC

AB = 3

Distance travelled by the balloon, OB = AB - OA

OB = 87283 - 29873

= 3

Distance travelled by the balloon = 58 3 m.

15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Solution:

Let the height of the tower be AD and the starting point of the car be at point B and after 6 seconds let the car be at point C. The angles of the depression of the car from the top A of the tower at point B and C are 30° and 60° respectively.

Distance travelled by the car from the starting point towards the tower in 6 seconds = BC

Distance travelled by the car after 6 seconds towards the tower = CDAB

We know that, speed = distancetime

The speed of the car is calculated using the distance BC and time = 64 seconds.

Using Speed and Distance CD, the time to reach foot can be calculated.

In ΔABD,

tan 30° = ADBDBDAD

131−3 = ADBD

BD = 3 ....(1)

In ΔACD,

tan 60° = ADCDCDAD

313 = ADCD

AD = 3 ....(2)

From equation (1) and (2)

BD = 3 × 3

BC + CD = CD [∵ BD = BC + CD]

BC = CD ....(3)

Distance travelled by the car from the starting point towards the tower in 6 seconds = BC

Speed of the car to cover distance BC in 6 seconds = DistanceTime

= BC6AC6

= 2CD6 [from (3)]

= CD3BD3

Speed of the car = CD3 m/s

Distance travelled by the car from point C, towards the tower = CDAC

Time to cover distance CD at the speed of CD3 m/s

Time = Distancespeed

= CDCD3

= CDAB × 3CD

=

The time taken by the car to reach the foot of the tower from point C is 3 seconds.