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From a height of 50 meters above sea level on a cliff, two ships are sighted due west. The angles of depression are 61° and 28°. How far apart are the ships?

Respuesta :

Answer:

The ships are 66 meters apart.

Step-by-step explanation:

For the sake of convenience, let us label ships A and B

As shown in the figure, the distances to the ships from right triangles.

The distance to the ship A is [tex]d_1[/tex] and it is given by

[tex]tan (61^o)= \dfrac{50}{d_1}[/tex]

[tex]d_1=\dfrac{50}{tan (61^o)}[/tex]

[tex]\boxed{d_1= 27.71m}[/tex]

And the distance to the ship B is [tex]d_2[/tex] and is given by

[tex]tan (28^o)= \dfrac{50}{d_2}[/tex]

[tex]d_2=\dfrac{50}{tan (28^o)}[/tex]

[tex]\boxed{ d_2=94.04m}[/tex]

Therefore, the distance [tex]d[/tex] between the ships A and B is

[tex]d= d_2-d_1=94.04-27.7\\\\\boxed{d=66m}[/tex]

In other words, the ships are 66 meters apart.

Ver imagen Poltergeist

Answer:

66.32 meters

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

In the right triangle ABC                  

Find the measure of side BC

[tex]tan(62^o)=\frac{50}{BC}[/tex] ---> TOA (opposite side divided by the adjacent side)

[tex]BC=\frac{50}{tan(61^o)}=27.72\ m[/tex]

step 2

In the right triangle ABD                  

Find the measure of side BD

[tex]tan(28^o)=\frac{50}{BD}[/tex] ---> TOA (opposite side divided by the adjacent side)

[tex]BD=\frac{50}{tan(28^o)}=94.04\ m[/tex]

step 3

How far apart are the ships?

[tex]CD=BD-BC[/tex]

substitute

[tex]CD=94.04-27.72=66.32\ m[/tex]

Ver imagen calculista