Question

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?

Answer 1

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 $d_1$ and it is given by

$tan (61^o)= \dfrac{50}{d_1}$

$d_1=\dfrac{50}{tan (61^o)}$

$\boxed{d_1= 27.71m}$

And the distance to the ship B is $d_2$ and is given by

$tan (28^o)= \dfrac{50}{d_2}$

$d_2=\dfrac{50}{tan (28^o)}$

$\boxed{ d_2=94.04m}$

Therefore, the distance $d$ between the ships A and B is

$d= d_2-d_1=94.04-27.7\\\\\boxed{d=66m}$

In other words, the ships are 66 meters apart.

Answer 2

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

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

$BC=\frac{50}{tan(61^o)}=27.72\ m$

step 2

In the right triangle ABD                  

Find the measure of side BD

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

$BD=\frac{50}{tan(28^o)}=94.04\ m$

step 3

How far apart are the ships?

$CD=BD-BC$

substitute

$CD=94.04-27.72=66.32\ m$