Respuesta :
Please refer to the figure attached for the diagram of this problem.
Steps needed to find the width of the field (CD):
First, we should note that angle d would be equal to 27 degrees because there are two parallel lines that are cut by a transversal. Furthermore, angle a would be equal to 5 degrees since we just need to subtract 27 from 32.
We then subtract 27 and 5 from 180 to get angle c. [tex]180-27-5=148[/tex]. Angle c is therefore 148 degrees.
Next, we need to find angle e which is just the supplementary of angle c. Angle e therefore measures [tex]180-148=32[/tex] degrees.
For the next step we use sine law to find the length of segment AC:
[tex] \frac{sin(32)}{800} = \frac{sin(90)}{AC} [/tex]
[tex]AC=1509.66[/tex]
Lastly, we need to utilize the sine law again to find the length of segment CD or the width of the field:
[tex] \frac{sin(27)}{1509.66} = \frac{sin(5)}{CD} [/tex]
[tex]CD=290[/tex]
ANSWER: The width of the field is 290 ft.
Steps needed to find the width of the field (CD):
First, we should note that angle d would be equal to 27 degrees because there are two parallel lines that are cut by a transversal. Furthermore, angle a would be equal to 5 degrees since we just need to subtract 27 from 32.
We then subtract 27 and 5 from 180 to get angle c. [tex]180-27-5=148[/tex]. Angle c is therefore 148 degrees.
Next, we need to find angle e which is just the supplementary of angle c. Angle e therefore measures [tex]180-148=32[/tex] degrees.
For the next step we use sine law to find the length of segment AC:
[tex] \frac{sin(32)}{800} = \frac{sin(90)}{AC} [/tex]
[tex]AC=1509.66[/tex]
Lastly, we need to utilize the sine law again to find the length of segment CD or the width of the field:
[tex] \frac{sin(27)}{1509.66} = \frac{sin(5)}{CD} [/tex]
[tex]CD=290[/tex]
ANSWER: The width of the field is 290 ft.
