Some Applications of Trigonometry MCQs

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Trigonometry is a branch of mathematics that deals with particular functions of the various angles and their application in different calculations. There are six types of functions used in trigonometry which can be termed sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec) and cosecant (csc).

Sample Questions

Ques. If the length of the shadow of a tree is decreasing then the angle of elevation is:

(a) Increasing

(b) Decreasing

(c) Remains the same

(d) None of the above

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Ans: (a) Increasing

Explanation: Observe the following figure AB is the height of the tree:

Observe the following figure AB is the height of the tree

As the length of the shadow decreases, it moves from point D to C towards the direction of the tree, the angle of elevation increases as it gets closer to the tree, just like in this figure it increases from 30° to 60°.

Ques. If the height of the building and distance from the building foot’s to a point is increased by 20%, then the angle of elevation on the top of the building:

(a) Increases

(b) Decreases

(c) Does not change

(d) None of the above

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Ans: (c) Does not change

Explanation: Let θ be the angle of deviation

tan θ = Height of building/Distance from the point

The formula shows that if we increase the values of both the height and distance equally, the angle of elevation will remain unchanged

Ques. If a tower 6m high casts a shadow of 2√3 m long on the ground, then the sun’s elevation is:

(a) 60°

(b) 45°

(c) 30°

(d) 90°

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Ans: (a) 60°

Explanation: As per the given question the following diagram will be formed, it explains the position and distance:

We know, θ is the angle of deviation.

Tan θ = Perpendicular / Base

Hence,

tan θ = 6/2√3

tan θ = √3

tan θ = tan 60°

⇒ θ = 60°

Ques. The angle of elevation of the top of a building 30 m high from the foot of another building in the same plane is 60°, and also the angle of elevation of the top of the second tower from the foot of the first tower is 30°, then the distance between the two buildings is:

(a) 10√3 m

(b) 15√3 m

(c) 12√3 m

(d) 36 m

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Ans: (a) 10√3 m

Explanation: As per the given question the following diagram will be formed PA being the length of

the first building and QB of the second building, AB is the distance between the two buildings.

Let the distance be x

Tan θ = Perpendicular / Base

Hence,

tan 60° = 30/x

√3 = 30/x

x = 30/√3

x = 10√3m

Ques. The angle formed by the line of sight with the horizontal when the point is below the horizontal level is called:

(a) Angle of elevation

(b) Angle of depression

(c) No such angle is formed

(d) None of the above

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Ans: (b) Angle of depression

Explanation: Analyse the figure:

The angle of depression is the angle formed by the line of sight with the horizontal when the point is below the horizontal level.

Ques. The angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level is called:

(a) Angle of elevation

(b) Angle of depression

(c) No such angle is formed

(d) None of the above

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Ans: (a) Angle of elevation

Explanation: Observe the following figure:

The angle of elevation is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level.

Ques. The angle of elevation of the top of a building from a point on the ground, which is 30 m away from the foot of the building, is 30°. The height of the building is:

(a) 10 m

(b) 30/√3 m

(c) √3/10 m

(d) 30 m

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Ans: (b) 30/√3 m

Explanation: Let x be the height of the building.

Let ‘m’ be a point 30 m away from the foot of the building.

Here, height is perpendicular, and the distance between points m and the foot of the building is the base.

The angle of elevation θ formed is 30°

Hence,

tan 30° = perpendicular/base = x/30

1/√3 = x/30

x = 30/√3

Ques. The line drawn from the eye of an observer to the point in the object viewed by the observer is said to be

(a) Angle of elevation

(b) Angle of depression

(c) Line of sight

(d) None of the above

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Ans: (c) Line of sight

Explanation: Observe the following figure

The Line of Sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.

Ques. The height or length of an object or the distance between two distant objects can be determined with the help of:

(a) Trigonometry angles

(b) Trigonometry ratios

(c) Trigonometry identities

(d) None of the above

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Ans: (b) Trigonometry ratios

Explanation: The height or length of an object, the distance between two distant objects, the angle of elevation or depression, or deviation can be determined with the help of trigonometry ratios.

some of the standard ratios used for applying trigonometry:

Sin θ = Perpendicular / Hypotenuse

Cos θ = Base / Hypotenuse

Tan θ = Perpendicular / Base

Cosec θ = 1 / sin =Hypotenuse / Perpendicular

Sec θ = 1 / cos = Hypotenuse / Base

Cot θ = 1 / tan = Base / Perpendicular

Ques. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. The height of the tower (in m) standing straight is:

(a) 15√3

(b) 10√3

(c) 12√3

(d) 20√3

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Ans: (a) 15√3

Explanation: Let the angle of elevation be θ and h be the height of the tower

We know,

tan θ (angle of elevation) = height of tower/its distance from the point

hence,

tan 60° = h/15

√3 = h/15

h = 15√3

Ques. When the shadow of a pole h meters high is √3h meters long, the angle of elevation of the Sun is

(a) 30°

(b) 60°

(c) 45°

(d) 15°

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Ans: (a) 30°

Explanation: Let AB be the pole and BC be its shadow.

Consider θ as the angle of elevation of the Sun.

In the right triangle ABC,

We know that,

tan θ (angle of elevation) = height of pole AB /its distance from the point BC

Hence,

tan θ = AB/BC = h/√3h = 1/√3

tan θ = tan 30°

θ = 30°

Ques. A ladder makes an angle of 60° with the ground when placed along a wall. If the foot of the ladder is 8 m away from the wall, the length of the ladder is

(a) 4 m

(b) 8 m

(c) 8√3 m

(d) 16 m

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Ans: (d) 16 m

Explanation:

Let AB be the wall, AC be the length of the ladder.

In right angled triangle ABC,

We know that,

Cos θ = Base / Hypotenuse

Hence,

cos 60° = BC/AC

½ = 8/AC

AC = 8 × 2 = 16

On solving, the length of the ladder comes out to 16 m.

Ques. If the height and length of a shadow of a tower are the same, then the angle of elevation of the Sun is

(a) 30°

(b) 60°

(c) 45°

(d) 15°

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Ans: (c) 45°

Explanation:

Let AB be the tower and BC be its shadow.

Consider the angle of elevation of the Sun as θ.

Given that,

AB = BC

We know that,

tan θ (angle of elevation) = height of tower AB /its distance from the point BC

Hence,

In right triangle ABC,

tan θ = AB/BC

tan θ = AB/AB {since AB = BC}

tan θ = 1

tan θ = tan 45°

θ = 45°

Ques. The angle of depression of an object on the ground, from the top of a 25 m high tower is 30°. The distance of the object from the base of the tower is

(a) 25√3 m

(b) 50√3 m

(c) 75√3 m

(d) 50 m

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Answer: (a) 25√3

Explanation:

Let the length of the tower be AB and BC be the distance of the object from the tower.

In the right triangle ABC,

We know that,

tan θ (angle of elevation) = height of pole AB /its distance from the point BC

Hence,

tan 30° = AB/BC

1/√3 = 25/BC

BC = 25√3 m

Therefore, the distance of the object from the base of the tower is 25√3 m.

Also Read:

Experimental Probability	Elementary Event	Sum of Probabilities
Geometric Probability	Sure Event	Probability Impossible Events
Complementary Events	Events in Probability	Empirical Probability
Equally Likely Events	Theoretical Probability	Class Mark
Cumulative Frequency Curve	Angle of Depression	Basic Proportionality Theorem

Some Applications of Trigonometry MCQs

Sample Questions

CBSE X Related Questions

1.
Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.

2.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

3.
A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

4.
In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

5.
Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$

6.
The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

Similar Mathematics Concepts

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Some Applications of Trigonometry MCQs

Sample Questions

CBSE X Related Questions

1.Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.

2.PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

3.A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

4.In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

5.Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$

6.The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.

2.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

3.
A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

4.
In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

5.
Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$

6.
The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.