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Trigonometric Ratios - JSS3 Mathematics Past Questions and Answers - page 1

1

What is the trigonometric ratio used to find the height of an object when given the angle of elevation and horizontal distance?

A

sin

B

cos

C

tan

correct option: d
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2

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?

A

20 meters

B

25 meters

C

16.32 meters

correct option: c
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3

Angle of depression is the angle formed when:

A

Looking up from horizontal to see an object

B

Looking down from horizontal to see an object

C

Looking horizontally at an object

correct option: b
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4

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?

A

19.32 meters

B

12.46 meters

C

10 meters

correct option: a
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5

If tan(θ) = 0.75, then θ is approximately:

A

36.87 degrees

B

48.59 degrees

C

37.43 degrees

correct option: c
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6

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?

A

13.45 meters

B

12.32 meters

C

12.98 meters

correct option: b
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7

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.

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8

An 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.

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9

An 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.

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10

An 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.

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