Some Applications of Trigonometry — Previous Year Questions (Class 10 Mathematics)
Trigonometry has numerous practical applications in surveying, navigation, and physics. This chapter covers angle of elevation, angle of depression, and so
TL;DR: Trigonometry has numerous practical applications in surveying, navigation, and physics. This chapter covers angle of elevation, angle of depression, a…
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Trigonometry has numerous practical applications in surveying, navigation, and physics. This chapter covers angle of elevation, angle of depression, and so
Some Applications of Trigonometry — Previous Year Questions with Solutions
Q (2023, 3 marks): A ladder of length 13 m is leaning against a wall. If the ladder makes an angle of 60 degrees with the ground, find the height of the wall it reaches.
Answer: Given:
- Length of ladder (hypotenuse) = 13 m
- Angle with ground = 60 degrees
Find: Height of wall
Solution:
Let height of wall = h
In right triangle:
sin(60 degrees) = h / 13
sqrt(3)/2 = h / 13
h = 13 * sqrt(3)/2
h = 13sqrt(3)/2 m
h ≈ 11.26 m
Answer: Height of wall = 13sqrt(3)/2 m or approximately 11.26 m
Q (2022, 3 marks): From the top of a 60 m high cliff, the angle of depression to a boat in the sea is 30 degrees. Find the horizontal distance of the boat from the cliff.
Answer: Given:
- Height of cliff = 60 m
- Angle of depression = 30 degrees
Find: Horizontal distance of boat
Solution:
Angle of depression = Angle of elevation from boat's perspective
Let distance = d
tan(30 degrees) = height / distance
1/sqrt(3) = 60 / d
d = 60 * sqrt(3)
d = 60sqrt(3) m
d ≈ 103.9 m
Answer: Horizontal distance = 60sqrt(3) m or approximately 103.9 m
Q (2023, 4 marks): Two poles of heights 10 m and 15 m stand vertically on level ground. If the distance between them is 12 m, find the angle between the line joining their tops and the ground.
Answer: Given:
- Height of pole 1 = 10 m
- Height of pole 2 = 15 m
- Distance between poles = 12 m
Find: Angle between line joining tops and ground
Solution:
Difference in heights = 15 - 10 = 5 m
Horizontal distance = 12 m
tan(angle) = 5 / 12
angle = arctan(5/12)
angle ≈ 22.62 degrees
Answer: The angle is approximately 22.62 degrees or arctan(5/12)
Q (2021, 4 marks): A person standing on the ground observes the top of a building at an angle of elevation of 45 degrees. If the person moves 20 m towards the building, the angle of elevation becomes 60 degrees. Find the height of the building.
Answer: Given:
- Initial angle of elevation = 45 degrees
- Angle after moving 20 m closer = 60 degrees
- Distance moved = 20 m
Find: Height of building
Solution:
Let height = h, initial distance = x
From first position: tan(45 degrees) = h/x => h = x
From second position: tan(60 degrees) = h/(x-20)
sqrt(3) = h/(x-20)
sqrt(3)(x-20) = h
Substituting h = x:
sqrt(3)(x-20) = x
sqrt(3)x - 20sqrt(3) = x
x(sqrt(3)-1) = 20sqrt(3)
x = 20sqrt(3)/(sqrt(3)-1)
Rationalizing: x = 20sqrt(3)(sqrt(3)+1)/((sqrt(3)-1)(sqrt(3)+1))
x = 20sqrt(3)(sqrt(3)+1)/2 = 10sqrt(3)(sqrt(3)+1)
x = 30 + 10sqrt(3)
Height h = 30 + 10sqrt(3) m ≈ 47.32 m
Answer: Height of building = (30 + 10sqrt(3)) m
Q (2022, 4 marks): From a point on the ground, the angle of elevation to the top of a tree is 60 degrees. From a point 50 m away from the first point (in the opposite direction), the angle of elevation is 30 degrees. Find the height of the tree.
Answer: Given:
- Angle of elevation from first point = 60 degrees
- Angle of elevation from second point = 30 degrees
- Distance between points = 50 m
Find: Height of tree
Solution:
Let height = h, distance from first point = x
From first point: tan(60 degrees) = h/x => sqrt(3) = h/x => h = x*sqrt(3)
From second point: tan(30 degrees) = h/(50+x)
1/sqrt(3) = h/(50+x)
(50+x)/sqrt(3) = h
Equating both expressions for h:
x*sqrt(3) = (50+x)/sqrt(3)
3x = 50 + x
2x = 50
x = 25
Height h = 25*sqrt(3) m ≈ 43.3 m
Answer: Height of tree = 25sqrt(3) m or approximately 43.3 m
Q (2021, 2 marks): The shadow of a vertical pole 30 m high is observed to be 30sqrt(3) m on level ground. Find the angle of elevation of the sun.
Answer: Given:
- Height of pole = 30 m
- Length of shadow = 30sqrt(3) m
Find: Angle of elevation of sun
Solution:
tan(angle) = height / shadow
tan(angle) = 30 / (30sqrt(3))
tan(angle) = 1/sqrt(3)
angle = 30 degrees
Answer: Angle of elevation of sun = 30 degrees
Frequently Asked Questions
What is the difference between angle of elevation and angle of depression?
Angle of elevation is measured upward from the horizontal to see an object above, while angle of depression is measured downward from the horizontal to see an object below.
What are the standard trigonometric values for 30, 45, and 60 degrees?
sin(30)=1/2, cos(30)=sqrt(3)/2, tan(30)=1/sqrt(3); sin(45)=cos(45)=1/sqrt(2), tan(45)=1; sin(60)=sqrt(3)/2, cos(60)=1/2, tan(60)=sqrt(3)
More Class 10 Mathematics PYQs
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