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NCERT Solutions for Class 10th Maths:Trigonometry applications





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° (see below figure).
NCERT Solutions for Class 10th Maths:Trigonometry applications
Solution

Let AB be the vertical pole AC be 20 m long rope tied to point C.
In ΔABC,
sin 30° = Perpendicular/Base=AB/AC
⇒ 1/2 = AB/20
⇒ AB = 20/2
⇒ AB = 10
The height of the pole is 10 m.
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
Let us draw the situation below
NCERT Solutions for Class 10th Maths:Trigonometry applications
Let AC be the broken part of the tree.
AB +AC is the tree height
In right ΔABC,
cos 30° = Base/Hypotenuse =BC/AC
⇒ √3/2 = 8/AC
⇒ AC = 16/√3
Also,
tan 30° = AB/BC
⇒ 1/√3 = AB/8
⇒ AB = 8/√3
Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3
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 :
Let us take small slide first
NCERT Solutions for Class 10th Maths:Trigonometry applications
AB is the height  and AC is the length of the slide
In right ΔABC,
sin 30° = Perpendicular/hypotenuse=AB/AC
⇒ 1/2 = 1.5/AC
⇒ AC = 3m
Now take the case of elder children slide
NCERT Solutions for Class 10th Maths:Trigonometry applications
PQ is the height
PR is the length of slide
In r ΔPQR,
sin 60° = Perpendicular/hypotenuse= PQ/PR
⇒ √3/2 = 3/PR
⇒ PR = 2√3 m
Hence, length of the slides are 3 m and 2√3 m respectively.

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:
The situation is depicted  in the below figure
NCERT Solutions for Class 10th Maths:Trigonometry applications
AB is height of the tower
BC =30 m (given)
In right ΔABC,
tan 30° = Perpendicular/base=AB/BC
⇒ 1/√3 = AB/30
⇒ AB = 10√3
Thus, the height of the tower is 10√3 m.
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:
The situation is depicted in the below figure
NCERT Solutions for Class 10th Maths:Trigonometry applications
Let BC be the height of the kite from the ground,
AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.
In  ΔABC,
sin 60° =Perpendicular/base= BC/AC
⇒ √3/2 = 60/AC
⇒ AC = 40√3 m
Thus, the length of the string from the ground is 40√3 m.
Question 6:
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building
Solution:
The situation is depicted in the below figure
NCERT Solutions for Class 10th Maths:Trigonometry applications
Given
DY=1.5m
Let the boy initially standing at point Y with inclination 30° and then he approaches the building to the point X with inclination 60°.
 so XY is the distance he walked towards the building.
Height of the building = AZ=30 m
also, XY = CD.
AB = AZ - BZ = (30 - 1.5) = 28.5 m
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 28.5/BD
⇒ BD = 28.5√3 m
also,
In right ΔABC,
tan 60° = AB/BC
⇒ √3 = 28.5/BC
⇒ BC = 28.5/√3 = 28.5√3/3 m
∴ XY = CD = BD - BC = (28.5√3 - 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.
Thus, the distance boy walked towards the building is 57/√3 m.
Question 7:
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution:
The situation is depicted in the below figure
NCERT Solutions for Class 10th Maths:Trigonometry applications
D is the point on Ground
BC be the 20 m high building..
Height of transmission tower = AB = AC - BC
In right ΔBCD,
tan 45° = BC/CD
⇒ 1 = 20/CD
⇒ CD = 20 m
also,
In right ΔACD,
tan 60° = AC/CD
⇒ √3 = AC/20
⇒ AC = 20√3 m
Height of transmission tower = AB = AC - BC = (20√3 - 20) m = 20(√3 - 1) m.
Question 8:
A statue, 1.6 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.
Solution:
The situation is depicted in the below figure
NCERT Solutions for Class 10th Maths:Trigonometry applications
Let AB be the height of statue.
D is the point on the ground from where the elevation is taken.
Height of pedestal = BC = AC - AB
In right ΔBCD,
tan 45° = BC/CD
⇒ 1 = BC/CD
⇒ BC = CD.
also,
In right ΔACD,
tan 60° = AC/CD
⇒ √3 = AB+BC/CD
⇒ √3CD = 1.6 m + BC
⇒ √3BC = 1.6 m + BC
⇒ √3BC - BC = 1.6 m
⇒ BC(√3-1) = 1.6 m
⇒ BC = 1.6/(√3-1) m
⇒ BC = 0.8(√3+1) m
Thus, the height of the pedestal is 0.8(√3+1) m.
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