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SOME APPLICATIONS OF TRIGONOMETRY - Q&A

EXERCISE 9.1

1. 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).

Solution:
Let the height of the vertical pole be AB and the length of the rope be AC.
We are given:
Length of rope (Hypotenuse), AC = 20 m
Angle made by rope with ground, ∠C = 30°
We need to find the height of the pole, AB (Opposite side).

In right-angled triangle ABC, we use the sine ratio:
sin C = Opposite / Hypotenuse
sin 30° = AB / AC
We know that sin 30° = 1/2.
So, 1/2 = AB / 20
AB = 20 / 2
AB = 10 m

Answer: The height of the pole is 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:
Let the original height of the tree be the sum of the upright part and the broken part.
Let the upright part be AB and the broken part be AC (which forms the hypotenuse).
The top of the tree touches the ground at point C.
Given:
Distance from foot of tree (B) to point C = 8 m.
Angle of elevation, ∠C = 30°.

Step 1: Find the height of the upright part (AB).
In ΔABC:
tan 30° = Opposite / Adjacent = AB / BC
1/√3 = AB / 8
AB = 8/√3 m

Step 2: Find the length of the broken part (AC).
cos 30° = Adjacent / Hypotenuse = BC / AC
√3/2 = 8 / AC
AC = (8 × 2) / √3 = 16/√3 m

Step 3: Calculate total height of the tree.
Total Height = AB + AC
= 8/√3 + 16/√3
= 24/√3
Rationalize the denominator by multiplying by √3/√3:
= (24√3) / 3
= 8√3 m

Answer: The height of the tree is 8√3 m.


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:
Case 1: For younger children
Height (AB) = 1.5 m
Angle (∠C) = 30°
Length of slide (AC) = ?
Using sine ratio:
sin 30° = AB / AC
1/2 = 1.5 / AC
AC = 1.5 × 2 = 3 m

Case 2: For elder children
Height (PQ) = 3 m
Angle (∠R) = 60°
Length of slide (PR) = ?
Using sine ratio:
sin 60° = PQ / PR
√3/2 = 3 / PR
PR = (3 × 2) / √3
PR = 6 / √3
Rationalize: (6√3) / 3 = 2√3 m

Answer: The length of the slide for younger children is 3 m and for elder children is 2√3 m.


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:
Let the height of the tower be h meters.
Distance from foot of tower = 30 m.
Angle of elevation = 30°.
In the right-angled triangle formed:
tan 30° = Height / Distance
1/√3 = h / 30
h = 30 / √3
Rationalize the denominator:
h = (30√3) / 3
h = 10√3 m

Answer: The height of the tower is 10√3 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:
Height of kite (Opposite) = 60 m
Angle of inclination = 60°
Length of string (Hypotenuse) = L
Using sine ratio:
sin 60° = Height / String Length
√3/2 = 60 / L
L = (60 × 2) / √3
L = 120 / √3
Rationalize:
L = (120√3) / 3
L = 40√3 m

Answer: The length of the string is 40√3 m.


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:
Total height of building = 30 m.
Height of boy = 1.5 m.
Height of building above boy's eye level (h) = 30 - 1.5 = 28.5 m.

Let the initial distance of the boy from the building be 'x'.
Let the distance he walked be 'd'.
Let the remaining distance to the building be 'y'. So, x = d + y.

First Position (Angle 30°):
tan 30° = h / x
1/√3 = 28.5 / x
x = 28.5√3 m

Second Position (Angle 60°):
tan 60° = h / y
√3 = 28.5 / y
y = 28.5 / √3
Rationalize y: y = (28.5√3) / 3 = 9.5√3 m

Distance walked (d):
d = x - y
d = 28.5√3 - 9.5√3
d = 19√3 m

Answer: The distance he walked towards the building is 19√3 m.


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 be BC = 20 m.
Let the height of the transmission tower be AB = h meters.
Let the point on the ground be D, at distance x from the building (CD = x).

For triangle BCD (Angle 45° to bottom of tower):
tan 45° = BC / CD
1 = 20 / x
x = 20 m

For triangle ACD (Angle 60° to top of tower):
Total height AC = 20 + h
tan 60° = AC / CD
√3 = (20 + h) / 20
20√3 = 20 + h
h = 20√3 - 20
h = 20(√3 - 1) m

Answer: The height of the tower is 20(√3 - 1) 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 h meters.
Height of statue = 1.6 m.
Total height = h + 1.6 m.
Let the distance of the point from the base be x meters.

For the top of the pedestal (Angle 45°):
tan 45° = Height of pedestal / Base
1 = h / x
x = h

For the top of the statue (Angle 60°):
tan 60° = Total height / Base
√3 = (h + 1.6) / x
Since x = h, we substitute:
√3 = (h + 1.6) / h
h√3 = h + 1.6
h√3 - h = 1.6
h(√3 - 1) = 1.6
h = 1.6 / (√3 - 1)
Rationalize the denominator by multiplying 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

Answer: The height of the pedestal is 0.8(√3 + 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 (AB) = 50 m.
Let the height of the building (CD) = h m.
Let the distance between their feet (BD) = x m.

From foot of building to top of tower (Angle 60°):
In ΔABD:
tan 60° = AB / BD
√3 = 50 / x
x = 50 / √3 m

From foot of tower to top of building (Angle 30°):
In ΔCDB:
tan 30° = CD / BD
1/√3 = h / x
h = x / √3
Substitute the value of x:
h = (50 / √3) / √3
h = 50 / 3
h = 16 2/3 m or 16.67 m

Answer: The height of the building is 16 2/3 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 the height of both poles be 'h'.
Width of road = 80 m.
Let the point be P. Let distance from one pole be 'x', then distance from the other pole is '80 - x'.

Pole 1 (Angle 60°):
tan 60° = h / x
√3 = h / x
h = x√3 --- (Equation 1)

Pole 2 (Angle 30°):
tan 30° = h / (80 - x)
1/√3 = h / (80 - x)
h = (80 - x) / √3 --- (Equation 2)

Equating both values of h:
x√3 = (80 - x) / √3
x√3 * √3 = 80 - x
3x = 80 - x
4x = 80
x = 20 m

Distance from first pole = 20 m.
Distance from second pole = 80 - 20 = 60 m.
Height h = x√3 = 20√3 m.

Answer: The height of the poles is 20√3 m, and the distances of the point from the poles are 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 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). Find the height of the tower and the width of the canal.

Solution:
Let the height of the tower AB = h.
Let the width of the canal BC = x.
The second point D is 20 m away from C, so DB = 20 + x.

In ΔABC (Angle 60°):
tan 60° = h / x
√3 = h / x
h = x√3 --- (1)

In ΔABD (Angle 30°):
tan 30° = h / (20 + x)
1/√3 = h / (20 + x)
h = (20 + x) / √3 --- (2)

Equate h from (1) and (2):
x√3 = (20 + x) / √3
3x = 20 + x
2x = 20
x = 10 m (Width of canal)

Height h = x√3 = 10√3 m.

Answer: The height of the tower is 10√3 m and the width of the canal is 10 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 height of building AB = 7 m.
Let the cable tower be CE.
From top of building A, angle of elevation to top of tower (C) is 60° and depression to foot (E) is 45°.
Draw a horizontal line AD from A to the tower.
Then, in rectangle ABED, BE = AD and DE = AB = 7 m.
Let the height of the part of the tower above the building (CD) be h.

Step 1: Find horizontal distance AD.
In ΔADE (using angle of depression 45°, which equals alternate interior angle ∠AED):
Or simply in right triangle formed by depression:
tan 45° = DE / AD
1 = 7 / AD
AD = 7 m

Step 2: Find height h.
In ΔADC:
tan 60° = CD / AD
√3 = h / 7
h = 7√3 m

Step 3: Total height of tower.
Total height = DE + CD = 7 + 7√3 = 7(1 + √3) m.

Answer: The height of the tower is 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:
Height of lighthouse (AB) = 75 m.
Let the two ships be at points C and D.
Angle of depression for C = 45° (closer ship).
Angle of depression for D = 30° (farther ship).
Let distance of C from lighthouse be y and distance between ships (CD) be x.

For Ship C (Angle 45°):
tan 45° = 75 / y
1 = 75 / y
y = 75 m

For Ship D (Angle 30°):
tan 30° = 75 / (y + x)
1/√3 = 75 / (75 + x)
75 + x = 75√3
x = 75√3 - 75
x = 75(√3 - 1) m

Answer: The distance between the two ships 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:
Height of balloon from ground = 88.2 m.
Height of girl = 1.2 m.
Effective height relative to girl's eyes (h) = 88.2 - 1.2 = 87 m.
Let the initial horizontal distance be x and the distance travelled be d. Final distance = x + d.

First Position (Angle 60°):
tan 60° = h / x
√3 = 87 / x
x = 87 / √3
x = 29√3 m

Second Position (Angle 30°):
tan 30° = h / (x + d)
1/√3 = 87 / (x + d)
x + d = 87√3

Calculate d:
d = 87√3 - x
d = 87√3 - 29√3
d = 58√3 m

Answer: The distance travelled by the balloon is 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 h.
Let the initial position of the car be A (angle 30°) and position after 6 seconds be B (angle 60°).
Let the foot of the tower be C.
Let distance BC = x and distance AB = y.
Speed is uniform, so time is proportional to distance.

In ΔBCT (Angle 60°):
tan 60° = h / x
√3 = h / x
h = x√3

In ΔACT (Angle 30°):
tan 30° = h / (x + y)
1/√3 = h / (x + y)
x + y = h√3
Substitute h = x√3:
x + y = (x√3)√3
x + y = 3x
y = 2x

Time Calculation:
The car travels distance y (which is 2x) in 6 seconds.
Therefore, it travels distance x (which is half of y) in half the time.
Time = 6 / 2 = 3 seconds.

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

Quick Navigation:
Quick Review Flashcards - Click to flip and test your knowledge!
Question
The line drawn from the eye of an observer to the point in the object being viewed is called the _____.
Answer
Line of sight
Question
What is the angle of elevation?
Answer
The angle formed by the line of sight with the horizontal when the point viewed is above the horizontal level.
Question
How is the angle of depression defined?
Answer
The angle formed by the line of sight with the horizontal when the point viewed is below the horizontal level.
Question
In viewing an object at an angle of elevation, what action must the observer perform with their head?
Answer
Raise their head.
Question
In viewing an object at an angle of depression, what action must the observer perform with their head?
Answer
Lower their head.
Question
What three pieces of information are needed to calculate a minar's height $CD$ if an observer of height $AE$ stands at distance $DE$?
Answer
The distance $DE$, the angle of elevation $\angle BAC$, and the height of the student $AE$.
Question
If an observer's height is considered, how is the total height of an object calculated using the segment $BC$ found by trigonometry?
Answer
Total height = $BC$ + observer's height.
Question
Which trigonometric ratio is most commonly used to relate the height of an object to its distance from the observer?
Answer
$\tan A$ (Tangent)
Question
Why is $\tan A$ or $\cot A$ typically chosen to find the height of a tower when the distance from the foot is known?
Answer
These ratios involve both the opposite side (height) and the adjacent side (distance).
Question
A tower stands vertically; from a point 15 m away, the angle of elevation to the top is $60^{\circ}$. What is the height?
Answer
$15\sqrt{3}$ m
Question
What is the exact value of $\tan 60^{\circ}$?
Answer
$\sqrt{3}$
Question
An electrician repairs a 5 m pole by reaching a point 1.3 m below the top. What is the height $BD$ of the target point?
Answer
3.7 m
Question
To find the length of a ladder (hypotenuse) given the vertical height and an angle of $60^{\circ}$, which ratio is used?
Answer
$\sin 60^{\circ}$
Question
What is the numerical value for $\sin 60^{\circ}$?
Answer
$\frac{\sqrt{3}}{2}$
Question
The electrician's ladder makes a $60^{\circ}$ angle to reach a 3.7 m height. What is the approximate length of the ladder?
Answer
4.28 m
Question
How far from the pole should the foot of the ladder be placed if height is 3.7 m and the angle is $60^{\circ}$?
Answer
Approx. 2.14 m
Question
An observer 1.5 m tall is 28.5 m from a chimney with an angle of elevation of $45^{\circ}$. Find the total height of the chimney.
Answer
30 m
Question
What is the exact value of $\tan 45^{\circ}$?
Answer
1
Question
In Example 4, if the angle of elevation of a 10 m building is $30^{\circ}$ from point $P$, what is the distance from $P$ to the building?
Answer
$10\sqrt{3}$ m
Question
A flagstaff $x$ is on top of a 10 m building. If the angle of elevation to the top of the flagstaff is $45^{\circ}$, what is the total height $(10 + x)$?
Answer
The total height equals the distance from the point $P$ to the building.
Question
What is the approximate length of the flagstaff in Example 4 using $\sqrt{3} = 1.732$?
Answer
7.32 m
Question
In Example 5, what happens to the length of a shadow as the Sun's altitude (angle of elevation) decreases?
Answer
The shadow becomes longer.
Question
If the shadow of a tower is 40 m longer at $30^{\circ}$ than at $60^{\circ}$, what is the height of the tower?
Answer
$20\sqrt{3}$ m
Question
In problems involving depression, why are the angle of depression and the angle of elevation of the same line of sight equal?
Answer
They are alternate angles formed by the transversal and parallel horizontal lines.
Question
From a multi-storeyed building, the angles of depression of an 8 m building's top and bottom are $30^{\circ}$ and $45^{\circ}$. Calculate the taller building's height.
Answer
$4(3 + \sqrt{3})$ m
Question
A bridge is 3 m high. The angles of depression of opposite banks are $30^{\circ}$ and $45^{\circ}$. What is the width of the river?
Answer
$3(\sqrt{3} + 1)$ m
Question
What is the exact value of $\tan 30^{\circ}$?
Answer
$\frac{1}{\sqrt{3}}$
Question
Exercise 9.1 Q1: A 20 m rope is tied at $30^{\circ}$ to the ground. What is the height of the pole?
Answer
10 m
Question
Exercise 9.1 Q2: A tree breaks at an angle of $30^{\circ}$ with the ground, 8 m from the foot. What is the total height of the tree?
Answer
$8\sqrt{3}$ m
Question
Exercise 9.1 Q4: Find the tower height if the angle of elevation from 30 m away is $30^{\circ}$.
Answer
$10\sqrt{3}$ m
Question
Exercise 9.1 Q5: A kite is at 60 m height with string at $60^{\circ}$. Find the length of the string.
Answer
$40\sqrt{3}$ m
Question
Exercise 9.1 Q12: From a 7 m building, the angle of elevation to a tower's top is $60^{\circ}$ and depression to its foot is $45^{\circ}$. What is the tower's height?
Answer
$7(\sqrt{3} + 1)$ m
Question
Exercise 9.1 Q15: A car moves from an angle of depression of $30^{\circ}$ to $60^{\circ}$ in 6 seconds. How much longer to reach the tower?
Answer
3 seconds
Question
When solving for the hypotenuse given an opposite side and an angle, which ratio is most appropriate?
Answer
$\sin \theta$ (or $\csc \theta$)
Question
In height and distance problems, the ground level is typically assumed to be a _____ line.
Answer
Horizontal
Question
What is the approximate value of $\sqrt{3}$ often used in calculations?
Answer
1.732
Question
If the angle of elevation of the top of a tower is $45^{\circ}$, what is the ratio of the tower's height to the observer's distance?
Answer
1:1 (The height and distance are equal).
Question
In $\Delta ABC$ where $\angle B = 90^{\circ}$, what is the formula for $\cot A$?
Answer
$\frac{AB}{BC}$ (Adjacent / Opposite)
Question
Which two variables are linked by the $\tan$ ratio in a right-angled triangle representing a building and its shadow?
Answer
Height of the building and length of the shadow.
Question
In Example 6, what is the distance between the multi-storeyed building and the 8 m building?
Answer
$4(3 + \sqrt{3})$ m
Question
What is the relationship between $\tan \theta$ and $\cot \theta$?
Answer
They are reciprocals ($\tan \theta = \frac{1}{\cot \theta}$).
Question
If an object is located exactly at the horizontal level of the observer's eye, what is the angle of elevation?
Answer
$0^{\circ}$
Question
What is the name of the branch of mathematics used to determine the height or length of distant objects in this chapter?
Answer
Trigonometry
Question
Concept: Sun's Altitude
Answer
The angle of elevation of the Sun from a point on the ground.
Question
In right $\Delta APD$ with $\angle A = 30^{\circ}$ and height $PD = 3$, find the length of $AD$.
Answer
$3\sqrt{3}$
Question
In $\Delta ABC$, if $\tan 60^{\circ} = \frac{h}{x}$, express $h$ in terms of $x$.
Answer
$h = x\sqrt{3}$
Question
Exercise 9.1 Q13: From a 75 m lighthouse, ships are seen at depression angles of $30^{\circ}$ and $45^{\circ}$. Find the distance between them.
Answer
$75(\sqrt{3} - 1)$ m
Question
What value should be added to the result of a $\tan$ calculation if the eye level is above the ground?
Answer
The height of the observer.
Question
In Exercise 9.1 Q7, how is the height of the transmission tower calculated relative to the 20 m building?
Answer
Subtract the building height (20 m) from the total height found using the $60^{\circ}$ angle.
Question
If the angle of elevation increases as an observer walks toward an object, is the observer moving closer to or further from the foot?
Answer
Closer to the foot.
Question
In Exercise 9.1 Q10, two poles of equal height $h$ are 80 m apart. If a point is distance $x$ from one, what is the distance from the other?
Answer
$80 - x$
Question
What is the distance between two objects on the same line from a tower if their depression angles are $\alpha$ and $\beta$?
Answer
$h(\cot \alpha - \cot \beta)$ (assuming $\alpha < \beta$).
Question
In $\Delta BDC$, if $\angle D = 90^{\circ}$, $\angle C = 60^{\circ}$, and $BD = 3.7$, what is $BC$?
Answer
$\frac{7.4}{\sqrt{3}}$ (approx. 4.28)
Question
The summary states that the height or length of an object can be determined using _____.
Answer
Trigonometric ratios
Question
The angle of elevation of an object viewed is formed by the line of sight and the _____.
Answer
Horizontal level