An obsever is looking out to sea from the top of a building that is 30 m above sea leavel. He observera ship at an angle of depression of 10. How far is the ship from the foot of the bulding?
2 answers
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Let O represent the observer,
B = base of the building (vertically below the observer O), and
S = the observed ship.
OBS represents a right-triangle right-angled at B.
Angle of depression
=∠BSO
= 10°
Solve the right triangle:
OB = 30m
&BSO = 10°
tan(10°)=OB/BS = 30m / BS
Solve for BS
BS = 30m /tan(10°)
=30m / 0.17633 (approx.)
= 170 m (approx.)
Let O represent the observer,
B = base of the building (vertically below the observer O), and
S = the observed ship.
OBS represents a right-triangle right-angled at B.
Angle of depression
=∠BSO
= 10°
Solve the right triangle:
OB = 30m
&BSO = 10°
tan(10°)=OB/BS = 30m / BS
Solve for BS
BS = 30m /tan(10°)
=30m / 0.17633 (approx.)
= 170 m (approx.)
3 answers