Answer:
22.8°
Explanation:
Given conditions:
When light beams to the east it is inclined at 3 degrees upwards
When the light beams points to the north it is inclined 10 degrees upwards
Determine the maximum angle of elevation of the light beam
let slope of the line ( m ) = tan ∅
In x-direction ( eastwards ) = df / dx = tan 3°
In y-direction ( north ) = df / dy = tan 10°
| νf | = | < tan 3° , tan 10° > |
therefore | νf | = [tex]\sqrt{(tan 3)^2 + (tan10)^2}[/tex]
| νf | = [tex]\sqrt{0.02032 +0.42037 }[/tex] = √ 0.44069 = 0.42037
hence the maximum angle of elevation = tan^-1 ( 0.42037 ) = 0.39794 rad
to degrees = 22.8°
The maximum angle of elevation of the light beam will be 22.8°. An elevation is its altitude beyond or below a specified reference point, It is calculated in degrees.
An elevation is its height above or below a fixed reference point, It is measured in degrees.
The given data in the problem is;
Angle inclined when light beams to the east = 3° upwards
Angle inclined light beams point to the north = 10° upwards
m is the slope of the line = tan ∅
The maximum angle of elevation of the light beam is found as;
The angle inclined in the x and y-direction;
In x-direction ( eastwards ) = df / dx = tan 3°
In y-direction ( north ) = df / dy = tan 10°
The resultant angle of both the above angle is;
[tex]\rm v_f = \sqrt{(tan3^0)^2 +(tan 10^0)^2 } \\\\ \rm v_f = \sqrt{(0.020320)^ +(0.42037) } \\\\ \rm v_f = \sqrt{0.44069 \\\\[/tex]
[tex]\rm v_f = 0.42037[/tex]
[tex]\rm \theta= tan^-1 (v_f) \\\\ \rm \theta= tan^-1 (0.42037) \\\\ \rm \theta= 22.8 ^0[/tex]
Hence the maximum angle of elevation of the light beam will be 22.8°.
To learn more about the elevation refer to the link;
https://brainly.com/question/481548
