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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Answer:
66170 mm
Step-by-step explanation:
0.06617 km * 1000 km/m * 1000 m/mm = 66170 mm
Answer:
to find the answer, you need to find the measure of the angle on the other side of the 142. and since it's a straight line, those two angles measures are =to 180 (they're supplementary angles). so the measure of that angle is 38 degrees. to find the measure of the remaining angle, you subtract the angle measures you already know from 180 (because the sum of interior angles in a triangle is 180.) to, the answer is:
180-90-38= 52 degrees or A
A- (6,-3) B- (9,-3) C- (9,-7)
Answer:
256
Step-by-step explanation:
64*4
(60*4)+(4*4)
240+16
256
you could also take a different approach--
64=4^3
4^3*4=4^4
4^4=16^2=256
