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From a plane flying due east at 265 m above sea level, the angles of depression of two ships sailing due east measure 35 degrees and 25 degrees. How far apart are the two ships?

Respuesta :

Answer:

189.8 m

Step-by-step explanation:

See the diagram attached.

Now, the plane is at B and the two ships are at C and at D.

So, angle of depression of C from B is ∠ OBC = ∠ ACB = 35°

Again, the angle of depression of D from B is ∠ OBD = ∠ ADB = 25° ,

Now, from the right triangle Δ ACB,  

[tex]\tan 35 = \frac{AB}{AC} = \frac{265}{AC}[/tex]

⇒ [tex]AC = \frac{265}{\tan 35} = 378.5[/tex] m.

Similarly, from the right triangle Δ ADB,

[tex]\tan 25 = \frac{AB}{AD} = \frac{265}{AD}[/tex]

⇒ [tex]AD = \frac{265}{\tan 25} = 568.3[/tex] m.

Hence, the distance between the ships = CD = AD - AC = 568.3 - 378.5 = 189.8 m (Approx.) (Answer)

Ver imagen rani01654

The distance apart is the two ships is 189.8 meters.

Given

A plane flying due east at 265 m above sea level, the angles of depression of two ships sailing due east measure 35 degrees and 25 degrees.

What is the angle of depression?

The angle of elevation, the angle of depression is the angle at which you would draw a line from a point on a horizontal line to intersect another point that falls below the line.

The plane is at B and the two ships are at C and at D.

Then,

The angle of depression of C from B is ∠ OBC = ∠ ACB = 35°.

And the angle of depression of D from B is ∠ OBD = ∠ ADB = 25°.

From the right triangle Δ ACB,

[tex]\rm Tan35= \dfrac{AB}{AC}\\\\ AC= \dfrac{265}{Tan35}\\\\AC = \dfrac{265}{0.70}\\\\AC =378.57[/tex]

Again in the right triangle Δ ADB,

[tex]\rm Tan 25=\dfrac{AB}{AD}\\\\AD =\dfrac{265}{Tan25}\\\\ AD = \dfrac{265}{0.46}\\\\AD =569.30[/tex]

Therefore,

The distance apart are the two ships is;

=  568.3 - 378.5 = 189.8 m

Hence, the distance apart is the two ships is 189.8 meters.

To know more about Angle of Depression click the link given below.

https://brainly.com/question/4790507

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