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NCERT Solutions for Class 10 Maths Chapter 9 Applications of Trigonometry

NCERT Solutions for Class 10 Maths Chapter 9 Applications of Trigonometry (Heights and Distances) covers solving real-life problems involving heights and distances using the angle of elevation and angle of depression.

Overview of Topics & Subtopics (Click to Jump)

1. Angle of Elevation and Angle of Depression

Short Note:

• Angle of Elevation: the angle formed by the line of sight with the horizontal, when the object being viewed is above the horizontal level, i.e., the observer looks upward.

• Angle of Depression: the angle formed by the line of sight with the horizontal, when the object being viewed is below the horizontal level, i.e., the observer looks downward.

2. Solving Heights and Distances Problems

Short Note:

• Such problems are solved by drawing a suitable right-triangle diagram based on the given situation, and applying appropriate trigonometric ratios (sin, cos, tan) of the known angle to find unknown heights or distances.

NCERT Textbook Exercise Questions

A tower stands vertically on the ground. From a point 20 m away from the foot of the tower, the angle of elevation of the top is 60°. Find the height of the tower.
Answer: tan 60° = height/20 ⇒ √3 = h/20 ⇒ h = 20√3 m ≈ 34.64 m.
A kite is flying at a height of 60 m above the ground. The string makes an angle of 60° with the horizontal. Find the length of the string, assuming no slack.
Answer: sin 60° = 60/L ⇒ √3/2 = 60/L ⇒ L = 120/√3 = 40√3 m.
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°. Find the height of the tower.
Answer: Horizontal distance = 7 m (from angle of depression, tan45°=7/d ⇒ d=7). For the tower's top: tan60° = (H−7)/7 ⇒ √3 = (H−7)/7 ⇒ H = 7 + 7√3 m ≈ 19.12 m.
Two poles of equal height stand on either side of a road 80 m wide. From a point between them, the angles of elevation of the tops are 60° and 30°. Find the height of the poles and the distances of the point from the poles.
Answer: Let height = h and distances be x and (80−x). tan60° = h/x and tan30° = h/(80−x). From these, h = x√3 and h = (80−x)/√3. Solving simultaneously gives x = 20 m, h = 20√3 m ≈ 34.64 m, and the other distance is 60 m.
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Class 10 › Mathematics › Applications of Trigonometry
Chapter notesRead the chapterPractice MCQsCheck what you rememberChapter test10 questions on one pageOnline testTest 1, 2 and 3, one question at a timePYQPrevious year questionsStudy materialPDF notes
Topics in this chapter: Angle of Elevation and Angle of DepressionSolving Heights and Distances Problems