Answer:
The 13.2 meters from that point on the ocean floor to the wreck.
Explanation:
As given
A ship's sonar finds that the angle of depression to a ship wrack on the bottom of the ocean is 12.5°.
If a point on the ocean floor is 60 meters.
Now by using the trigonometric identity.
![tan\theta = (Perpendicular)/(Base)](https://img.qammunity.org/2019/formulas/mathematics/high-school/h5afpp5hy2wlsipxotjdvp4u61pb1c6998.png)
As shown in the figure given below.
Perpendicular = CB
Base = AC = 60 meters
![\theta = 12.5^(\circ)](https://img.qammunity.org/2019/formulas/mathematics/high-school/dk3wn3chr8bt1uyucdrwefeyd5h1lgav43.png)
Put in the identity
![tan\ 12.5^(\circ) = (CB)/(AC)](https://img.qammunity.org/2019/formulas/mathematics/high-school/72es6pl8kvv2rxqssp7fpgmgozdg0x35yx.png)
![tan\ 12.5^(\circ) = (CB)/(60)](https://img.qammunity.org/2019/formulas/mathematics/high-school/2tmyv9umkbglsxyofqmic9pd3ygulflgw2.png)
![tan\ 12.5^(\circ) = 0.22](https://img.qammunity.org/2019/formulas/mathematics/high-school/5akeafsch9h3qjjr4vzvm46zlgwf9myvkm.png)
![tan\ 12.5^(\circ) = 0.22](https://img.qammunity.org/2019/formulas/mathematics/high-school/5akeafsch9h3qjjr4vzvm46zlgwf9myvkm.png)
![0.22= (CB)/(60)](https://img.qammunity.org/2019/formulas/mathematics/high-school/7scnk4mf3eh5xrxxcinhv0d78r8lp087qa.png)
CB = 60 × 0.22
CB = 13.2 meters
Therefore the 13.2 meters from that point on the ocean floor to the wreck.