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 = [tex]\dfrac{\textrm perpendicular}{\textrm hypotenuse}[/tex]
Or, Sin Ф = [tex]\dfrac{\textrm x}{\textrm e}[/tex]
Or, Sin 45° = [tex]\dfrac{\textrm x meters}{\textrm 1.2 meters}[/tex]
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 = [tex]\dfrac{\textrm perpendicular}{\textrm base}[/tex]
Or, Tan Ф = [tex]\dfrac{\textrm H}{\textrm y}[/tex]
Or, Tan 22° = [tex]\dfrac{\textrm 10,000 feet}{\textrm y feet}[/tex]
Or, 0.4040 = [tex]\dfrac{\textrm 10,000 feet}{\textrm y feet}[/tex]
Or, y = [tex]\dfrac{\textrm 10,000 feet}{\textrm 0.4040}[/tex]
∴ 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
