An observer stands 30 meters above sea level an... - JSS3 Mathematics Trigonometric Ratios Question
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.
Please share this, thanks:
Add your answer
No responses