Question

Help math homework both questions

Answer 1

Answer:

Q-12 The vertical length of the top part of elevator to ground is 0.8484 meters .

Q -14 The glide runway path is 24,752.47 feet .

Step-by-step explanation:

Q - 12

Given as :

The diagonal length of the elevator = e = 1.2 meters

The vertical length of the top part of elevator to ground = x meters

The angle made by elevator with ground = Ф = 45°

Now, According to figure

Sin angle = $\dfrac{\textrm perpendicular}{\textrm hypotenuse}$

Or, Sin Ф = $\dfrac{\textrm x}{\textrm e}$

Or, Sin 45° =  $\dfrac{\textrm x meters}{\textrm 1.2 meters}$

Or, 0.707 × 1.2 meters = x

∴ x =  0.8484 meters

So, The vertical length of the top part of elevator to ground = x = 0.8484 meters

Hence,The vertical length of the top part of elevator to ground is 0.8484 meters .  Answer

Q-14

Given as:

The altitude of decent of plane = H = 10,000 feet

The angle of descend =  Ф = 22°

Let the glide runway path = y feet

Now, According to question

Tan angle = $\dfrac{\textrm perpendicular}{\textrm base}$

Or, Tan Ф = $\dfrac{\textrm H}{\textrm y}$

Or, Tan 22° =  $\dfrac{\textrm 10,000 feet}{\textrm y feet}$

Or, 0.4040 =  $\dfrac{\textrm 10,000 feet}{\textrm y feet}$

Or, y =  $\dfrac{\textrm 10,000 feet}{\textrm 0.4040}$

∴   y = 24,752.47

So,The glide runway path = y = 24,752.47 feet

Hence, The glide runway path is 24,752.47 feet . Answer