Some applications of trigonometry class 10 notes chapter 9 are available with NCERT Solutions at . The notes are typically designed by the mathematics masters at the top-notch online education portal keeping in mind the updated pattern and guidelines by the CBSE board. These quick notes on CBSE class 10 maths will help you to significantly refine your trigonometric skills as well get to the core of the topic in no time.
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Examples of Class 10 Chapter 9 – Some Applications of Trigonometry
Under this section, you will find the solved questions of chapter 9 – Some Applications of Trigonometry from Class 10 Maths textbook along with answer keys. These solutions are available for free PDF download from the given link at official. Let’s get started with the solved Maths Class 10 chapter 9 questions.
Example: A villager is climbing a coconut tree using a 20 m long rope, which is tightly tied from the top of a vertical pole to the ground. Evaluate the height of the coconut tree, if the angle formed by the rope with the ground level is 30°. You can find the illustration below.
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Solution: Given that length of the rope named AC = 20m
Angle formed is ∠ACB = 30°
Let the height AB of the coconut tree be h (meters)
With that, in right ▲ABC,
sin 30° = AB/AC
½= h/20 (since sin 30° = ½)
H = 20/2 = 10metres
Therefore, the height of the coconut tree is 10m
Example: The facilities department of a housing society plans to put two slides for the kids to play in a park. For the kids below the age of 5 years, they want to have a slide whose top is at a height of 1.5 m, and is disposed at an angle of 30° to the ground, while for older children, they prefer to have a steep slide based at a height of 3 m, and inclined at an angle of 60° to the ground. Find out the length of the slide in both the cases?
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Solution:
Let the length of the slide for children below 5years of age x and the length of the slide for older children be y.
Given that: AF = 1.5 m and BC = 3m
∠FEA = 30°and ∠CDB = 60°
In right ▲FAE, sin 30°= AF/EF = 1.5/x
½ = 1.5/x
Thus x= 3m
In right ▲CBD, sin 60°= BC/CD = 3/y
√3/2 = 3/y
Thus, y = 3*2 / √3 = 2√3m
Therefore the length of the slide for children below 5years of age is 3m and the length of the slide for older children is 2√3m.
Example: An air balloon is flying above the ground at a height of 60 m. The string joined to the balloon is temporarily tied to a point on the ground. The inclination of the string with the ground level is 60°. What will be the length of the string, supposing that there is no slack in the string?
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Solution: Given that: AB = 60m and ∠ACB = 60°
Let AC be the length of the string
Then, in the right ▲ABC, sin 30°= AB/AC
= √3/2 = 60/AC
AC= 60*2/√3/ * √3/√3
= 120* √3 / 3
= 40√3
Therefore, the total length of the string is 40√3m