Class 10 Maths Objective Questions Chapter 9 Applications Of Trigonometry
CBSE Class 10 Maths Chapter 9 – Applications of Trigonometry Objective Questions
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 9-Applications of Trigonometry Objective Questions for the students to prepare for the exams. Students will learn the applications of trigonometry with real-life 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 sub-topics it covers.
We have compiled here some 20 multiple choice questions covering the below-given 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)
Solution: The given situation is represented by the figure below
The given situation can be represented by figure above
∴tan60°=DC/BC⇒BC=DC/tan60°= 60/ √3 = 20 √3 m
⇒AB=BC×tan30°=20 √3 × (1/√3 m) = 20m
The above figure represents the situation given in the question
⇒DC=20m, which is the required distance.
Height of the pole that is above man’s height = 2 √3 – √3 = √3m
Solution: In ΔABC, taking tangent of ∠C, we have,
Hence, the height of the tower is 10√3 metres
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.
Hence, the height of the tower is 30 m.
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.
Hence, the length of the ladder should be m.
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.
Hence, the height of the pole is 10 m.
The given situation is represented by the figure above:
Hence, the height of the chimney = AC = AB + BC
Solution: The given situation can be represented by the Δ below
The above figure represents the situation aptly
Solution: The given situation can be represented by the figure below
Solution: The situation can be represented by the figure below:
In the given right-angled triangle:
The situation can be represented by the figure above
The value of (M 2 −N 2 ) / (MN) 0.5
Solution: M 2 -N 2 = (Tan A+ Sin A + Tan A –Sin A) (Tan A +Sin A – Tan A+ Sin A)
and (MN) 0.5 = (tan 2 A−sin 2 A) 0.5
Let the height of tower A be = AB = H.
tan30° = AB/AC = H/AC ……………………………. 1
tan60° = CD/AC = h/AC………………………………….2
The above figure represents the situation given in question
Solution: Let C and D be the objects and CD be the distance between the objects.
Above you have access to the PDF link to download multiple choice questions from the chapter 9-Applications 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 StudySolver 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.