Answer:
Option A) 183 meters
Explanation:
see the attached figure to better understand the problem
step 1
Find the distance BC
we know that
In the right triangle ABC
----> adjacent side divided by the hypotenuse
Remember that
![cos(60\°)=1/2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jqm0ihrzpuf8s3s83boul7csfx48kpg2aw.png)
substitute the values
![1/2=(a)/(500)](https://img.qammunity.org/2020/formulas/mathematics/high-school/szmk588k8exc9xippmvyok1g8n31l8lx0v.png)
![a=250\ m](https://img.qammunity.org/2020/formulas/mathematics/high-school/96fdkdvtyofp1b82sq9tkkjramipqy7b8i.png)
step 2
Find the distance AC
In the right triangle ABC
----> opposite side divided by the hypotenuse
Remember that
![sin(60\°)=(√(3))/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/z3ja9tm18kmt86s2bg5u42l10mcg7wvpbe.png)
substitute the values
![(√(3))/(2)=(b)/(500)](https://img.qammunity.org/2020/formulas/mathematics/high-school/rumvvqc1csdo89vjhwcjpb24jguxqxhaq8.png)
![b=250√(3)= 433\ m](https://img.qammunity.org/2020/formulas/mathematics/high-school/ofe1t0zch97cefr1e6iwnb1tqjod599xl5.png)
step 3
Find the distance AC+CB
![433+250=683\ m](https://img.qammunity.org/2020/formulas/mathematics/high-school/d6tndcokpebd7i7q3ocm79f2zs1v1y0fcv.png)
Subtract the distance AB from 683 m
![683-500=183\ m](https://img.qammunity.org/2020/formulas/mathematics/high-school/gjt9i2cu9gfrm2n844ybcyp8fp9h8xx19a.png)