SOME APPLICATIONS OF TRIGONOMETRY

EXERCISE 9.1

Q1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11 (refer book)).

Solution:
In the right-angled triangle ABC:

• AB is the height of the pole (vertical side).

• AC is the rope = 20 m (hypotenuse).

• ∠C = 30° is the angle between the rope and the ground BC.

From trigonometry, in a right-angled triangle:
sin (angle) = Opposite side / Hypotenuse

Here,
sin 30° = AB / AC
sin 30° = AB / 20

We know sin 30° = 1/2.
So,
1/2 = AB / 20

Multiplying both sides by 20:
AB = 20 × (1/2)
AB = 10 m

Therefore, the height of the pole is 10 metres.

Answer: 10m

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Q2: 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:
Let the original height of the tree be H metres.
Let the point where the tree broke be A and the foot of the tree be B. The broken top AC touches the ground at C.

In the right-angled triangle ABC:

• BC = 8 m (distance from foot to point where top touches ground)

• ∠C = 30°

• AC = broken part of the tree (hypotenuse)

• AB = unbroken standing part of the tree (vertical side)

From trigonometry:
cos 30° = BC / AC
√3 / 2 = 8 / AC
AC = 8 × (2 / √3)
AC = 16 / √3 m

Now, the standing part of the tree AB is:
AB = AC × sin 30°
AB = (16 / √3) × (1/2)
AB = 8 / √3 m

Total height of the tree = AB + AC
= (8 / √3) + (16 / √3)
= 24 / √3 m

Simplifying:
= (24√3) / 3
= 8√3 m

Therefore, the height of the tree is 8√3 m (approximately 13.86 m).

Answer: 8√3 m

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Q3: 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 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution:

In a right triangle, sin(angle) = opposite side / hypotenuse.
Here, “opposite side” = height of the slide, and “hypotenuse” = length of the slide.

1. For the small slide:
height = 1.5 m, angle = 30°
sin 30° = 1/2
length = height / sin 30° = 1.5 / (1/2) = 3 m

2. For the steep slide:
height = 3 m, angle = 60°
sin 60° = √3 / 2
length = height / sin 60° = 3 / (√3 / 2) = 6 / √3 = 2√3 m ≈ 3.46 m

Answer:
Small slide length = 3 m
Steep slide length = 2√3 m (approximately 3.46 m)

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Q4: 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:
Let the height of the tower be h metres.
Distance from point to foot of tower = 30 m (base of the right triangle).
Angle of elevation = 30°.

Using the tangent formula:
tan 30° = h / 30

We know tan 30° = 1 / √3.
So,
1 / √3 = h / 30

Multiply both sides by 30:
h = 30 / √3

Simplify:
h = (30√3) / 3
h = 10√3 m

Therefore, the height of the tower is 10√3 m (approximately 17.32 m).

Answer: 10√3 m

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Q5: 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:
Let the length of the string be L metres.
Height of kite = 60 m (opposite side)
Angle of inclination = 60°

Using the sine formula:
sin 60° = opposite side / hypotenuse
√3 / 2 = 60 / L

Multiply both sides by L:
L × (√3 / 2) = 60

Multiply both sides by 2:
L × √3 = 120

Divide by √3:
L = 120 / √3

Simplify:
L = (120√3) / 3
L = 40√3 m

Therefore, the length of the string is 40√3 m (approximately 69.28 m).

Answer: 40√3 m

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Q6: 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:

Let the initial distance from the boy to the building be x metres.
The height from the boy’s eyes to the top of the building = 30 – 1.5 = 28.5 m.

Step 1: Before walking (angle = 30°)
tan 30° = (28.5) / x
1 / √3 = 28.5 / x
x = 28.5√3

Step 2: After walking (angle = 60°)
Let the new distance from the boy to the building be y metres.
tan 60° = (28.5) / y
√3 = 28.5 / y
y = 28.5 / √3

Step 3: Distance walked
Distance walked = x – y
= 28.5√3 – (28.5 / √3)
= 28.5 (√3)–(1/√3)(√3) – (1 / √3)
= 28.5 (3–1)/√3(3 – 1) / √3
= 28.5 × (2 / √3)
= (57 / √3)
= (57√3) / 3
= 19√3 m

Final Answer: Distance walked = 19√3 m (approximately 32.91 m).

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Q7: 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 tower be h metres.
Let the horizontal distance from the point on the ground to the building be x metres.

Step 1: Using the 45° angle (bottom of tower = top of building)
tan 45° = 20 / x
1 = 20 / x
x = 20 m

Step 2: Using the 60° angle (top of tower)
Total height from ground to top of tower = 20 + h
tan 60° = (20 + h) / x
√3 = (20 + h) / 20

Multiply both sides by 20:
20√3 = 20 + h

Step 3: Find h
h = 20√3 – 20
h = 20(√3 – 1) m

Final Answer:
Height of tower = 20(√3 – 1) m ≈ 14.64 m

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Q8: 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 h metres.
Let the horizontal distance from the point on the ground to the pedestal be x metres.

Step 1: From the 45° angle (top of pedestal)
tan 45° = h / x
1 = h / x
x = h

Step 2: From the 60° angle (top of statue)
Total height = h + 1.6
tan 60° = (h + 1.6) / x
√3 = (h + 1.6) / h

Multiply both sides by h:
h√3 = h + 1.6

Step 3: Solve for h
h√3 – h = 1.6
h(√3 – 1) = 1.6
h = 1.6 / (√3 – 1)

Multiply numerator and denominator by (√3 + 1):
h = [1.6(√3 + 1)] / [(√3 – 1)(√3 + 1)]
h = [1.6(√3 + 1)] / (3 – 1)
h = [1.6(√3 + 1)] / 2
h = 0.8(√3 + 1) m

Final Answer:
Height of pedestal = 0.8(√3 + 1) m ≈ 2.18 m

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Q9: 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 building be h metres.
Let the horizontal distance between the building and the tower be x metres.
Height of the tower = 50 m.

From the foot of the tower looking at the top of the building (angle 30°):
tan 30° = h / x
1 / √3 = h / x
So, x = h√3.

From the foot of the building looking at the top of the tower (angle 60°):
tan 60° = 50 / x
√3 = 50 / x
So, x = 50 / √3.

Equate the two expressions for x:
h√3 = 50 / √3
Multiply both sides by √3:
3h = 50
h = 50 / 3 m.

Therefore, the height of the building = 50/3 m = 16 2/3 m.

Answer: 16 (2/3) m.

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Q10: 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 the height of each pole be h metres.
Let the point on the road be at a distance x metres from the pole that subtends 60°.
Then the distance from the other pole (which subtends 30°) is 80 − x metres.

Using right-triangle trigonometry:

For the pole at angle 60°:
tan 60° = h / x
√3 = h / x
h = x√3

For the pole at angle 30°:
tan 30° = h / (80 − x)
1/√3 = h / (80 − x)
h = (80 − x)/√3

Equate the two expressions for h:
x√3 = (80 − x)/√3
3x = 80 − x
4x = 80
x = 20

So the distances from the point are:
20 m to the pole seen at 60°, and 60 m to the pole seen at 30°.

Height of each pole:
h = x√3 = 20√3 m (approximately 34.64 m)

Answer:
Height of each pole = 20√3 m.
Distances from the point = 20 m and 60 m (to the poles with elevation angles 60° and 30°, respectively).

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Q11: 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 joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12 (refer book)). Find the height of the tower and the width of the canal.

Solution:

Let BA = h be the height of the tower and BC = w be the width of the canal.
Point C is directly opposite the tower; point D is 20 m farther from the tower along the same line, so BD = BC + CD = w + 20.

At C:
tan 60° = h / w
√3 = h / w ⇒ h = √3 · w

At D:
tan 30° = h / (w + 20)
1/√3 = h / (w + 20) ⇒ h = (w + 20) / √3

Equating h from the two relations:
√3 · w = (w + 20) / √3
3w = w + 20
2w = 20 ⇒ w = 10 m

Height of the tower:
h = √3 · w = √3 · 10 = 10√3 m ≈ 17.32 m

Answer:
Width of the canal = 10 m.
Height of the tower = 10√3 m (approximately 17.32 m).

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Q12: 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 building’s top be point A, its foot be directly below A. Let the cable tower stand on the same horizontal ground, with foot T and top P. Let the horizontal distance AT(ground) = x and the tower’s height = H.

1. Angle of depression to the foot T is 45°.
tan 45° = opposite / adjacent = (vertical drop from A to ground) / x
1 = 7 / x ⇒ x = 7 m.

2. Angle of elevation to the top P is 60°.
tan 60° = (vertical rise from A to P) / x = (H − 7) / 7
√3 = (H − 7) / 7 ⇒ H − 7 = 7√3 ⇒ H = 7(√3 + 1) m.

Therefore, the height of the cable tower is 7(√3 + 1) m ≈ 19.12 m.

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Q13: 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 lighthouse height be 75 m.
Let the nearer ship (greater angle of depression 45°) be at distance x from the lighthouse base, and the farther ship (angle of depression 30°) be at distance y from the base. Angles of depression equal the corresponding angles of elevation.

At the nearer ship (45°):
tan 45° = 75 / x
1 = 75 / x ⇒ x = 75 m

At the farther ship (30°):
tan 30° = 75 / y
1/√3 = 75 / y ⇒ y = 75√3 m

Distance between the ships:
y − x = 75√3 − 75 = 75(√3 − 1) m ≈ 54.9 m

Answer: 75(√3 − 1) metres (approximately 54.9 m).

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Q14: 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 (refer book)). Find the distance travelled by the balloon during the interval.

Solution:

Let the height of the girl’s eyes above ground be 1.2 m and the balloon’s height be 88.2 m.
Vertical difference between the balloon and the girl’s eyes = 88.2 − 1.2 = 87 m.

Let the horizontal distance from the girl to the balloon when the angle is 60° be x1, and when the angle is 30° be x2.

For angle 60°:
tan 60° = vertical difference / x1
√3 = 87 / x1 ⇒ x1 = 87 / √3

For angle 30°:
tan 30° = vertical difference / x2
1/√3 = 87 / x2 ⇒ x2 = 87 √3

Distance travelled by the balloon = x2 − x1
= 87 √3 − 87 / √3
= 87 (√3 − 1/√3)
= 87 · (2 / √3)
= 174 / √3
= 58 √3 m

Approximate value:
58 √3 ≈ 58 × 1.73205 ≈ 100.46 m

Answer:
The balloon travelled 58√3 metres, approximately 100.46 m.

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Q15: 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 h.
Let the horizontal distance of the car from the foot of the tower when the angle of depression is 30° be x.

From the geometry of depression = elevation,
for angle 30°: tan 30° = h / x ⇒ 1/√3 = h / x ⇒ x = √3 · h.

Six seconds later the angle of depression is 60°. Let the distance travelled in these 6 seconds be d. Then the new distance from the tower is x − d.

For angle 60°: tan 60° = h / (x − d) ⇒ √3 = h / (x − d) ⇒ x − d = h / √3.

So d = x − h / √3 = √3·h − h / √3 = (2h) / √3.

Speed of the car = distance travelled / time = d / 6 = (2h / √3) / 6 = h / (3√3).

Distance remaining to the foot from the second position = x − d = h / √3.

Time to reach the foot from this second position = (remaining distance) / (speed)
= (h / √3) / (h / (3√3)) = 3 seconds.

Answer: 3 seconds.

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