Trigonometric Ratios - JSS3 Mathematics Past Questions and Answers - page 1
What is the trigonometric ratio used to find the height of an object when given the angle of elevation and horizontal distance?
sin
cos
tan
An observer looks up at an angle of 40 degrees to the top of a tree. If the observer is 20 meters away from the base of the tree, what is the height of the tree?
20 meters
25 meters
16.32 meters
Angle of depression is the angle formed when:
Looking up from horizontal to see an object
Looking down from horizontal to see an object
Looking horizontally at an object
An observer stands 15 meters above the ground and looks down at an angle of 25 degrees to see the top of a building. If the horizontal distance from the observer to the building is 30 meters, what is the height of the building?
19.32 meters
12.46 meters
10 meters
If tan(θ) = 0.75, then θ is approximately:
36.87 degrees
48.59 degrees
37.43 degrees
An observer stands 50 meters away from the base of a cliff. If the angle of elevation to the top of the cliff is 15 degrees, what is the height of the cliff?
13.45 meters
12.32 meters
12.98 meters
An observer stands 40 meters away from a building and looks up at an angle of elevation of 35 degrees. Find the height of the building.
Given:
Horizontal distance (adjacent side, adj) = 40 meters
Angle of elevation (θ) = 35 degrees
To find the height (opposite side, opp):
tan(35∘)=opp40
opp=40⋅tan(35∘)
Using a calculator: tan(35∘)≈0.7002
opp≈40⋅0.7002=28.01 meters
So, the height of the building is approximately 28.01 meters.
Users' Answers & CommentsAn observer stands on top of a hill and looks down at an angle of depression of 20 degrees to see a road directly below. If the height of the hill is 60 meters, how far is the road from the base of the hill?
Given:
Height of the hill (opposite side, opp) = 60 meters
Angle of depression (θ) = 20 degrees
To find the horizontal distance (adjacent side, adj):
tan(20∘)=60adj
adj=60
tan(20∘)
adj= tan(20 ∘ )60
Using a calculator:
tan(20∘ )≈0.3640
adj≈60/0.3640=164.84 meters
So, the road is approximately 164.84 meters away from the base of the hill.
Users' Answers & CommentsAn observer stands on the ground and looks up at an angle of elevation of 25 degrees to see the top of a tower. If the tower is 50 meters tall, how far is the observer from the base of the tower?
Given:
Angle of elevation (θ) = 25 degrees
Height of the tower (opposite side, opp) = 50 meters
To find the horizontal distance (adjacent side, adj):
tan(25∘)=50/adj
adj=50
tan(25∘)50
Using a calculator:
tan(25∘)≈0.4663
adj≈ 0.4663/50=107.19 meters
So, the observer is approximately 107.19 meters away from the base of the tower.
Users' Answers & CommentsAn observer stands 30 meters above sea level and looks down at an angle of depression of 10 degrees to see a ship. If the horizontal distance from the observer to the ship is 200 meters, how far is the ship from sea level?
Given:
Height above sea level (opposite side, opp) = 30 meters
Angle of depression (θ) = 10 degrees
Horizontal distance (adjacent side, adj) = 200 meters
To find the vertical distance (height of the ship, opposite side, opp):
tan(10∘)=30/200
opp=200⋅tan(10 ∘ )
Using a calculator:
tan(10∘)≈0.1763
opp≈200⋅0.1763=35.26 meters
So, the ship is approximately 35.26 meters below sea level.
Users' Answers & Comments