Answer:
sorry you out of luck man
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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A - 3b = 4
a = b -2
(b - 2) - 3b = 4
b - 2 - 3b = 4
-2b = 4 + 2
-2b = 6
b = 6/-2
b = -3
a = b - 2
a = -3 -2
a = -5
to check: a = -5 ; b = -3 ⇒ (-5,-3)
a - 3b = 4
-5 - 3(-3) = 4
-5 + 9 = 4
4 = 4
