PQ is tangent to the circle at C. In the circle, m (The figure is not drawn to scale.)
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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 =
Or, Sin Ф =
Or, Sin 45° =
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 =
Or, Tan Ф =
Or, Tan 22° =
Or, 0.4040 =
Or, y =
∴ 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