The axis of a light in a lighthouse is tilted. When the light points east, it is inclined upward at 4 degree(s). When it points north, it is inclined upward at 6 degree(s). What is its maximum angle of elevation? (Hint: The maximum angle of elevation of plane of the beam above the horizontal plane is the same as the angle between the normal to the plane of the beam and the normal to the horizontal plane.)
6 degrees.
No it is more than 6. The beam is at 0 degrees when it points WNW and ESE and maximum at NNW, minimum at SSE.
its not 6 ! i know that
It will be the max of a sine wave which solves: \[ a*\sin(b+0) = 4\] \[ a*\sin(b+90) = 6\] \[ a*\sin(b+180) = -4\] \[ a*\sin(b+270) = -6\]
its none of them !
\[\sin(b) = (2/3)\sin(b+\pi/2)\] \[3\sin(b) = 2 \cos(b)\] \[b= Arctan(2/3)\] = 33.7 degrees \[a = 4/\sin(33.7) = 2\sqrt{13}\]
still none !
How bout 6.959 degrees
nope
uh well how bout 10.11 degrees? and if not what is the correct answer
nope ! it didnt work either
what is the numerical value for that !?
Well heres how to solve it I think. We construct the equation for the plane of the beam: z = ax + by (we place the plane of the beam intersecting the horizontal plane at 0.0.0) We know that the partial derivitaves in the x and y direction are respectively: sin 6 sin 4 so the equaiton for the plane of the beam is z = xsin6 + ysin4 The coefficients of a plane equation are the components of the normal vector so, as the problem hint implies, we calculate the angle between this normal vector and the normal to the horizontal. Beam plane normal : sin6 sin4 -1 Horizontal normal : 0 0 1 The dot product for two vectors A and B is calculated 2 ways. x1*x2 + y1*y2 + z1*z2 => (sin6 * 0 + sin4 * 0 + 1) |A||B|cosθ Setting these equal to solve for the angle" |A||B|cosθ = 1 So we calculate 1/sin26+sin24+1−−−−−−−−−−−−−√ So: θ=cos−1(1/(sin26+sin24+1−−−−−−−−−−−−−√)) which is 10.11
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