This chapter 9 of CBSE Class 10 Maths, known as, “ Some applications of trigonometry” from the textbook deals with how trigonometry helps to find the height and distance of different objects without any measurement. In earlier days, astronomers used trigonometry to calculate the distance from the earth to the planets and stars. Trigonometry is also mostly used in navigation and geography to locate the position in relation to the latitude and longitude. Keeping in mind the latest modification for the CBSE Exam Pattern, we have compiled here the CBSE Class 10 Maths Chapter 9Applications of Trigonometry Objective Questions for the students to prepare for the exams. Students will learn the applications of trigonometry with reallife examples in a better way with the help of the MCQs we have compiled.
Find the CBSE Class 10 Maths Objective Questions and the list of subtopics it covers.
List of SubTopics Covered In Chapter 9
We have compiled here some 20 multiple choice questions covering the belowgiven topics from the chapter.
9.1 Applications of Trigonometry (6 MCQs From The Given Topic)
9.2 Introduction (4 MCQs From The Topic)
9.3 Heights and Distances (10 MCQs Listed From The Topic)
Download Free CBSE Class 10 Maths Chapter 9Applications of Trigonometry Objective Questions PDF
Applications of Trigonometry

 A Technician has to repair a light on a pole of height 10 m. She needs to reach a point 1 m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use which, when inclined at an angle of 60^{∘} to the ground, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder?
Answer: (D) 6√3 m
Solution: The given situation is represented by the figure below
DB – DC = CB
⇒BC=9msin60°=BCAC⇒AC=BCsin60°=
∴height of ladder should be 6√3 m.

 A statue, 2 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.
Answer: (B)2(√3 1)
Solution:
CD = 2m
Let BC = x
AB = x (using tan 45∘)
In ΔABD
BD = AB √3 = x √3
We know CD = BD – BC = 2m
X(√3 – 1)
X = 2(√3 1 ) , which is height.

 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 60 m high, find the height of the building.
 30m
 40m
 20m
 10m
 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 60 m high, find the height of the building.
Answer: (C) 20m
Solution:
The given situation can be represented by figure above
∴tan60°=DC/BC⇒BC=DC/tan60°= 60/ √3 = 20 √3 m
tan30°=AB/BC
⇒AB=BC×tan30°=20 √3 × (1/√3 m) = 20m
Thus, height of building is 20m

 A TV tower stands vertically on a bank of a canal, with a height of 10 √3 m . 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 on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the distance between the opposite bank of the canal and the point with 30° angle of elevation.
 30m
 20m
 45m
 35m
 A TV tower stands vertically on a bank of a canal, with a height of 10 √3 m . 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 on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the distance between the opposite bank of the canal and the point with 30° angle of elevation.
Answer: (B) 20m
Solution:
The above figure represents the situation given in the question
tan60°=AB/ BC
⇒AB=BCtan60°= BC √3……….(1)
⇒BC=AB/tan60°=AB/ tan60°
tan30°=AB/BD=AB/ (CD+BC)
⇒DC+BC=AB/tan30°=AB √3
⇒DC=AB
⇒DC=20m, which is the required distance.

 As observed from the top of a 150 m high lighthouse from the sealevel, 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.
Answer: (B) 150 (√3 – 1)
Solution:
Here Lighthouse BC = 150 m
In ΔBDC,
BD = BC = 150m (using tan 45°)
In ΔABC,
AB = BC √3 (using tan 30°)
AB= 150√3
Hence distance AD = 150 (√3 – 1)

 An observer √3m tall is 3 m away from the pole 2√3 high. What is the angle of elevation of the top?
 60°
 30°
 45°
 90°
 An observer √3m tall is 3 m away from the pole 2√3 high. What is the angle of elevation of the top?
Answer: (B) 30°
Solution:
Concept: 1 Mark
Application: 1 Mark
Height of the pole that is above man’s height = 2 √3 – √3 = √3m
Hence, AB = √3m
BC = 3m
Hence, tan C = AB/BC =
⟹ C, angle of elevation = 30°
Introduction

 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: In ΔABC, taking tangent of ∠C, we have,
tan C=AB/BC
⇒tan 30°=h/30
⇒\(\frac{1}{\sqrt{3}}\) = h/30
=10(1.732)
=17.32m= 10√3 m
Hence, the height of the tower is 10√3 metres

 An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from her eyes is 45°. What is the height of the tower?
 20m
 10m
 40m
 30m
 An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from her eyes is 45°. What is the height of the tower?
Answer: (D) 30m
Solution: Let AB be the tower of height h and CD be the observer of height 1.5 m at a distance of 28.5 m from the tower AB.
In ΔAED, we have
tan45°=h/28.5
⇒1=h/28.5
⇒h=28.5 m
∴h=AE+BE=AE+DC
= (28.5+1.5) m=30 m
Height of tower = h + 1.5
= 28.5 + 1.5
= 30 m
Hence, the height of the tower is 30 m.

 An electrician has to repair an electric fault on a pole of height 4 m. He needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use which when inclined at an angle of 60° to the horizontal would enable him to reach the required position?
Answer: (A) (9√3) / 5
Solution: Let AC be the electric pole of height 4 m. Let B be a point 1.3 m below the top A of the pole AC.
Then, BC = AC – AB = (4 – 1.3) m = 2.7 m
Let BD be the ladder inclined at an angle of 60° to the horizontal.
In ΔBCD, we have
sin60°=2.7/L
Hence, the length of the ladder should be m.

 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°.
 10m
 15m
 20m
 35m
 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°.
Answer: (A) 10m
Solution: Let AB be the vertical pole and CA be the 20 m long rope such that its one end is tied from the top of the vertical pole AB and the other end C is tied to a point C on the ground.
In ΔABC, we have
sin 30°=h/20
⇒1/2=h/20
⇒h=10 m
Hence, the height of the pole is 10 m.
Heights and Distances

 An observer 2.25 m tall is 42.75 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?
 40m
 50m
 45m
 35m
 An observer 2.25 m tall is 42.75 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?
Answer: (C) 45m
Solution:
The given situation is represented by the figure above:
In triangle ABE,
tan45°=AB/ EB
Also, EB=DC
∴tan45°=AB/ DC
⇒AB=DC × tan 45°
⇒AB=1×42.75
Hence, the height of the chimney = AC = AB + BC
We can observe that BC = ED.
Thus, AC = AB + ED
= 42.75 + 2.25
= 45 m.

 A tower stands vertically on the ground. From a point on the ground, which is 30 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 30°. Find the height of the tower.
Answer: (B) 10√3 m
Solution: The given situation can be represented by the Δ below
Now, tan30°=AB/ BC
∴Height of tower is 10√3 m.

 The angles of depression of the top and the bottom of a 10 m tall building from the top of a multistoreyed building are 30° and 45°, respectively. Find the height of the multistoreyed building.
Solution:
The above figure represents the situation aptly
ED is the height of the building.

 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 2m and is inclined at an angle of 30° to the ground. What should be the length of the slide?
 4
 2
 1.5
 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 2m and is inclined at an angle of 30° to the ground. What should be the length of the slide?
Answer: (A) 4
Solution: The given situation can be represented by the figure below
In rightangled triangle ABC,
sin∠ABC=AC/AB=1/2
⇒sin30°=2/AB⇒AB= 2/ (1/2)
⇒AB=4m
∴Length of the slide is 4m.

 A kite is flying at a height of 30 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: The situation can be represented by the figure below:
In the given rightangled triangle:
sin (∠ACB) =AB/AC
⇒sin60°=AB/AC
⇒AC=AB/sin60°=
∴ Length of the string is 20√3 m

 A vertical pole of 30 m is fixed on a tower. From a point on the level ground, the angles of elevation of the top and bottom of the pole is 60° and 45°. Find the height of the tower.
Answer: (B) 15 (√3 + 1)
Solution:
The situation can be represented by the figure above

 The value of tan A +sin A=M and tan A – sin A=N.
The value of (M^{2}−N^{2}) / (MN)^{ 0.5}


 4
 3
 2
 1

Answer: (A) 4
Solution: M^{2}N^{2} = (Tan A+ Sin A + Tan A –Sin A) (Tan A +Sin A – Tan A+ Sin A)
M^{2}N^{2} =4 tan A sin A
and (MN)^{ 0.5} = (tan^{2}A−sin^{2}A)^{ 0.5}

 Two towers A and B are standing at some distance apart. From the top of tower A, the angle of depression of the foot of tower B is found to be 30°. From the top of tower B, the angle of depression of the foot of tower A is found to be 60°. If the height of tower B is ‘h’ m then the height of tower A in terms of ‘h’ is _____ m
Answer: (B) h/3 m
Solution:
Let the height of tower A be = AB = H.
And the height of tower B = CD = h
In triangle ABC
tan30° = AB/AC = H/AC ……………………………. 1
In triangle ADC
tan60° = CD/AC = h/AC………………………………….2
Divide 1 by 2
We get tan30°/tan60° = H/h
H= h/3

 A 1.5 m tall boy is standing at some distance from a 31.5 m tall building. If he walks ’d’ m towards the building the angle of elevation of the top of the building changes from 30° to 60° . Find the length d. (Take √3 = 1.73)
 30.15 m
 38.33m
 22.91m
 34.55m
 A 1.5 m tall boy is standing at some distance from a 31.5 m tall building. If he walks ’d’ m towards the building the angle of elevation of the top of the building changes from 30° to 60° . Find the length d. (Take √3 = 1.73)
Answer: (D) 34.55m
Solution:
The above figure represents the situation given in question
∴distance moved by boy is 34.55 m

 The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be 45° and 30°. Find the distance between the two objects. (Take √3 = 1.73,)
 200m
 150m
 107.5m
 73.2m
 The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be 45° and 30°. Find the distance between the two objects. (Take √3 = 1.73,)
Answer: (D) 73.2 m
Solution: Let C and D be the objects and CD be the distance between the objects.
In ΔABC, tan 45° = AB/AC
AB=AC=100 m
In ΔABD, tan 30° = AB/AD
CD=AD−AC=173.2−100=73.2 metres
Above you have access to the PDF link to download multiple choice questions from the chapter 9Applications of Trigonometry, which has been categorised as per the topics from which it is taken. Solving these questions will help the students to score better in the board exams, because as per the latest exam pattern the question papers are more likely to include more objective questions.
Apart from these MCQs , students can also download other resources at BYJU’S like the NCERT solutions, previous years papers, syllabus, study notes, sample papers and important questions for all the classes to prepare more efficiently. Keep learning and stay tuned for further updates on CBSE.