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°
<u>Now, According to figure</u>
Sin angle = ![\dfrac{\textrm perpendicular}{\textrm hypotenuse}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctextrm%20perpendicular%7D%7B%5Ctextrm%20hypotenuse%7D)
Or, Sin Ф = ![\dfrac{\textrm x}{\textrm e}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctextrm%20x%7D%7B%5Ctextrm%20e%7D)
Or, Sin 45° = ![\dfrac{\textrm x meters}{\textrm 1.2 meters}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctextrm%20x%20meters%7D%7B%5Ctextrm%201.2%20meters%7D)
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
<u>Now, According to question</u>
Tan angle = ![\dfrac{\textrm perpendicular}{\textrm base}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctextrm%20perpendicular%7D%7B%5Ctextrm%20base%7D)
Or, Tan Ф = ![\dfrac{\textrm H}{\textrm y}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctextrm%20H%7D%7B%5Ctextrm%20y%7D)
Or, Tan 22° = ![\dfrac{\textrm 10,000 feet}{\textrm y feet}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctextrm%20%2010%2C000%20feet%7D%7B%5Ctextrm%20y%20feet%7D)
Or, 0.4040 = ![\dfrac{\textrm 10,000 feet}{\textrm y feet}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctextrm%20%2010%2C000%20feet%7D%7B%5Ctextrm%20y%20feet%7D)
Or, y = ![\dfrac{\textrm 10,000 feet}{\textrm 0.4040}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctextrm%20%2010%2C000%20feet%7D%7B%5Ctextrm%200.4040%7D)
∴ 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