Question

A shark is swimming 28 feet below sea level. If the angle of

Answer 1

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,  

$\tan 13 = \frac{AB}{AS} = \frac{28}{AS}$

⇒ $AS = \frac{28}{\tan 13} = 121.28$ feet.

Therefore, the horizontal distance between the boat and the shark is AS = 121.28 feet, (Answer)

Answer 2

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 = $\dfrac{\textrm Perpendicular}{\textrm base}$

or, Tan 13° =  $\dfrac{\textrm OA}{\textrm AB}$

Or, Tan 13° =  $\dfrac{\textrm 28}{\textrm x}$

Or, 0.2308 =  $\dfrac{\textrm 28}{\textrm x}$

Or, x =  $\frac{28}{0.2308}$

∴ 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