The angle of the plane when it rose from the ground is 64.8 degrees
<h3>Application of trigonometry identity</h3>
Given the following parameters from the question
Altitude of the airplane H = 500m
Horizontal distance from airport "d" = 235
Required
angle of elevation
According to the trigonometry identity
tan x = opposite/adjacent
tan x = H/d
tan x = 500/235
tan x = 2.1277
x = arctan(2.1277)
x = 64.8 degrees
The angle of the plane when it rose from the ground is 64.8 degrees
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X = 650 + 60t
The cheetah must arrive at the same place at the same time:
x = 90t
650 + 60t = 90t
30t = 650
t = 650/30 ≅ 21.7 s
It takes the cheetah a little more than 20 s running at 90 ft/s to reach the moving antelope, so assuming the antelope can maintain its speed of 60, it is safe.
V<span>ertical angles are equal.
We have 4x = 2x + 18.
Then, 2x = 18;
x = 18 </span>÷ 2;
x = 9 degrees;
The measure of angle DBE is 2 × 9 + 18 = 36 degrees.
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