Answer:
Part 21) The distance from Football Field to the Park is [tex]10.95\ miles[/tex]
Part 22)
a) The value of a is [tex]20\ units[/tex]
b) The value of b is [tex]20.10\ units[/tex]
Step-by-step explanation:
Part 21)
Let
A -----> Football Field
B -----> Park
C -----> Home
D -----> Library
x -----> the distance from Football Field to the Park
In the right triangle ABD
[tex]cos(A)=10/x[/tex] -----> equation A
In the right triangle ABC
[tex]cos(A)=x/12[/tex] ----> equation B
equate equation A and equation B
[tex]10/x=x/12[/tex]
solve for x
[tex]x^{2}=120\\ \\x=10.95\ miles[/tex]
Part 22)
see the attached figure with letter to better understand the problem
step 1
Find the value of a
In the right triangle ABD
[tex]tan(ABD)=a/200[/tex] ----> equation A
In the right triangle ADC
[tex]tan(DAC)=2/a[/tex] ----> equation B
remember that angle ABD is congruent with angle DAC
therefore
equate equation A and equation B
[tex]a/200=2/a[/tex]
solve for a
[tex]a^{2}=400\\ \\a=20\ units[/tex]
step 2
Find the value of b
In the right triangle ADC
Applying the Pythagoras Theorem
[tex]b^{2}=a^{2}+2^{2}[/tex]
substitute the value of a
[tex]b^{2}=20^{2}+2^{2}[/tex]
[tex]b^{2}=404[/tex]
[tex]b=20.10\ units[/tex]
