Answer :
QUESTION 1
The diagram to visualise this question is shown in the first picture below
We model this as a right angle triangle and we can use trigonometry ratio to find [tex]x[/tex] which is the distance between the shooter and the pier.
[tex]tan(x)= \frac{opposite}{adjacent} [/tex]
[tex]tan(70)= \frac{x}{20} [/tex]
[tex]x=20tan(70)[/tex]
[tex]x=54.94...[/tex] which rounded to 55 feet
QUESTION 2
Referring to the second diagram, we have
[tex]tan(x)= \frac{opposite}{adjacent} [/tex]
[tex]tan(50)= \frac{x}{45} [/tex]
[tex]x=45tan(50)[/tex]
[tex]x=53.6289....[/tex] which round to 54 feet
QUESTION 3
The length of [tex]x[/tex] is given
[tex]tan(x)= \frac{opposite}{adjacent} [/tex]
[tex]tan(65)= \frac{x}{100} [/tex]
[tex]x=100tan(65)[/tex]
[tex]x=214.45...[/tex] which round to 214
Answers:
Question 1 is C
Question 2 is C
Question 3 is [there is no option for 214 feet]
The diagram to visualise this question is shown in the first picture below
We model this as a right angle triangle and we can use trigonometry ratio to find [tex]x[/tex] which is the distance between the shooter and the pier.
[tex]tan(x)= \frac{opposite}{adjacent} [/tex]
[tex]tan(70)= \frac{x}{20} [/tex]
[tex]x=20tan(70)[/tex]
[tex]x=54.94...[/tex] which rounded to 55 feet
QUESTION 2
Referring to the second diagram, we have
[tex]tan(x)= \frac{opposite}{adjacent} [/tex]
[tex]tan(50)= \frac{x}{45} [/tex]
[tex]x=45tan(50)[/tex]
[tex]x=53.6289....[/tex] which round to 54 feet
QUESTION 3
The length of [tex]x[/tex] is given
[tex]tan(x)= \frac{opposite}{adjacent} [/tex]
[tex]tan(65)= \frac{x}{100} [/tex]
[tex]x=100tan(65)[/tex]
[tex]x=214.45...[/tex] which round to 214
Answers:
Question 1 is C
Question 2 is C
Question 3 is [there is no option for 214 feet]


