Answer:
The horizontal distance between boat and the shark is 121.28 feet
Step-by-step explanation:
Given as :
The Distance of the shark swimming below sea level = AB = 28 feet
The angle of depression from a boat on water to shark = 13°
Let The horizontal distance between shark and boat = OA = x
Now, Tan angle = [tex]\dfrac{\textrm Perpendicular}{\textrm base}[/tex]
or, Tan 13° = [tex]\dfrac{\textrm OA}{\textrm AB}[/tex]
Or, Tan 13° = [tex]\dfrac{\textrm 28}{\textrm x}[/tex]
Or, 0.2308 = [tex]\dfrac{\textrm 28}{\textrm x}[/tex]
Or, x = [tex]\frac{28}{0.2308}[/tex]
∴ x = 121.28 feet
So The horizontal distance = x = 121.28 feet
Hence The horizontal distance between boat and the shark is 121.28 feet Answer
Answer:
121.28 feet
Step-by-step explanation:
See the diagram attached.
Let B is the position of the boat and BA = 28 feet is the height of the water surface from the level of the shark.
If the position of the shark is at S then angle of depression from boat to the sherk i.e. ∠ OBS = 13° and we have ∠ BSA = ∠ OBS = 13° {Alternate angles}
Now, using trigonometry in right triangle Δ ABS we can write,
[tex]\tan 13 = \frac{AB}{AS} = \frac{28}{AS}[/tex]
⇒ [tex]AS = \frac{28}{\tan 13} = 121.28[/tex] feet.
Therefore, the horizontal distance between the boat and the shark is AS = 121.28 feet, (Answer)
