NCERT Solutions Maths Ch-9 Some Applications of Trigonometry for Class 10th
The NCERT Solutions for Class 10 Maths Chapter 9, Some Applications of Trigonometry, provide students with a comprehensive guide to solving real-world problems using trigonometric concepts. This chapter primarily focuses on the practical uses of trigonometry, including determining heights and distances in various situations without directly measuring them. Students learn how to apply trigonometric ratios such as sine, cosine, and tangent to calculate angles and lengths in right-angled triangles. The chapter covers examples like finding the height of a tower, distance of a point from an observer, and the angle of elevation or depression of an object. With step-by-step explanations, these solutions help students gain a clear understanding of trigonometry’s applications in daily life, aiding their preparation for exams and improving problem-solving skills.
NCERT Solutions Maths Ch-9 Some Applications of Trigonometry
We try to teach you all Questions in easy way. We solve all chapter wise sums of maths textbook. In every chapter include NCERT solutions. For solutions of Exercise 9.1 click on Tabs :
Question 1.
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°.
Solution:
Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution:
Using given instructions, draw a figure. Let AC be the broken part of the tree. Angle C = 30°
BC = 8 m
To Find: Height of the tree, which is AB
Total height of the tree is the sum of AB and AC i.e. AB+AC
In right ΔABC,
Using Cosine and tangent angles,
cos 30° = BC/AC
We know that, cos 30° = √3/2
√3/2 = 8/AC
AC = 16/√3 …(1)
Also,
tan 30° = AB/BC
1/√3 = AB/8
AB = 8/√3 ….(2)
Therefore, total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3 = 8√3 m.
Question 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 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
Question 4.
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:
Question 5.
A kite is flying at a height of 60 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:
Height of the kite from the ground, BC = 60 m
AC = Inclined length of the string from the ground and
A is the point where string of the kite is tied.
To Find: Length of the string from the ground i.e. the value of AC
From the above figure,
sin 60° = BC/AC
⇒ √3/2 = 60/AC
⇒ AC = 40√3 m
Thus, the length of the string from the ground is 40√3 m.