Question

PLEASE NEED THIS DONE AS FAST AS POSSIBLE!!!!!!!!!!!!!!!!! IT MUST BE CORRECT

Answer 1

Answer:

1)7.288 feet

2)11.6 feet

3)safe

4) 3.7 feet

5) ∅$= tan^{-1}(1.5) = 56.31 degrees$

6)∅1$= tan^{-1}(2) = 63.43 degrees$

  ∅2$= tan^{-1}(1.2) = 50.19 degrees$

Step-by-step explanation:

1) The door barn is rectangular in shape. The length is 9 feet and The angle between diagonal and side is 39 degrees.

Applying trigonometry,

tan(39) = $\frac{opposite}{adjacent} = \frac{s}{9}$ = 0.809

Thus, s= $(9)(0.809) = 7.288 feet$

2) Applying pythagoras theorm,

   $(Diagonal)^{2} = (9)^{2} + (7.288)^{2} =134.115$

  Diagonal length (d) = 11.58 feet. Nearest tenth place = 11.6 feet

3) The length of ladder is 14 foot and height from ground is 13.5 feet.

Applying trigonometry,

sin(∅) =  $\frac{opposite}{hypotenous} = \frac{13.5}{14}$ = 0.964

∅ = angle of elevation = $sin^{-1}(0.964)$ = 74.57 ≈ 75 degrees.

Thus tractor can climb safely.

4)Applying, pythgoras theorm,

$14^{2} = (13.5)^{2} + x^{2}$

x = $\sqrt{14^{2}-(13.5)^{2}} = 3.708$

Thus, ladder should be placed at distance 3.7 feet

5)Let angle of elevation be ∅.

  tan(∅) = $\frac{30}{20} = 1.5$

  ∅$= tan^{-1}(1.5) = 56.31 degrees$

6)After moving 5 feet closer to barn, Let angle of elevation for light near barn be ∅1 and for farther one be ∅2.

Thus,

tan(∅1) = $\frac{30}{15} = 2$

  ∅1$= tan^{-1}(2) = 63.43 degrees$

tan(∅2) = $\frac{30}{25} = 1.2$

  ∅2$= tan^{-1}(1.2) = 50.19 degrees$