Question

(PLEASE dont ignore, NEED help❗️❗️❗️)

Answer 1

Part 21) The distance from Football Field to the Park is
`10.95\ miles`

Part 22)

a) The value of a is
`20\ units`

b) The value of b is
`20.10\ units`

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

`cos(A)=10/x` -----> equation A

In the right triangle ABC

`cos(A)=x/12` ----> equation B

equate equation A and equation B

`10/x=x/12`

solve for x

`x^(2)=120\\ \\x=10.95\ miles`

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

`tan(ABD)=a/200` ----> equation A

In the right triangle ADC

`tan(DAC)=2/a` ----> equation B

remember that angle ABD is congruent with angle DAC

therefore

equate equation A and equation B

`a/200=2/a`

solve for a

`a^(2)=400\\ \\a=20\ units`

step 2

Find the value of b

In the right triangle ADC

Applying the Pythagoras Theorem

`b^(2)=a^(2)+2^(2)`

substitute the value of a

`b^(2)=20^(2)+2^(2)`

`b^(2)=404`

`b=20.10\ units`