Directions For Building Lighthouses
directions for Building Lighthouses
Tirgonometry Math Assignment?
5 questions
1. Circle with quadraalateral of sides 4, 4, 2, 3. in that order. Find the length of the diagonal from in between the 4s to in between 3 and 2
2. Tall building/ Mountain find the height with this information. from one point away from the top the angle is 21degrees and 34 seconds. 507 meters away the angle is 35 degrees 14 seconds.
3. boat is manned by a seasoned skipper, inexpeiereced man, and you. Aurgument between how for away a landmark is. the inexpierienced man says it is 3 nautical miles away and the skipper says it s twice as far. you are 5.25 miles away from a lighthouse and at an angle of 36 d. away from the landmark. it is 3.4 between the house and landmark.
4. Investigating noise src. 2 microphones 100m apart. one finds direction one measures direction. the second heres the sound 0.1 s later than the first. the line of sound made an angle of 60 d with the line of mics. find distance to sound src to each microphone
ask for q 5 im out of char.
2. h/d=tan(35 14)
h/(d+507)=tan(21 34)
solve for h
d=h/tan(35 14)
d=h/tan(21 34)-507
h=455 m
3. Break the distance to the rock into two pieces by drawing a line from the lighthouse perpendicular to a line from the boat to Big Rock
d2 is the segment from the boat to the perpendicular
d1 is the segment from the perpendicular to Big Rock
h is the length of the perpendicular
d2=5.25*cos(36)
4.247
h^2=5.25^2-4.247^2
d1=sqrt(3.4^2-h^2)
d1=+/-1.46
so the distance is either
5.637 or 2.82
you can't tell without another bearing and knowing the speed and direction of the boat between bearings.
4) For this we need the speed of sound. I will use a constant c to set up the equations
d1 is the distance from the source to mic 1 and d2 is the distance from the source to mic 2
d2=d1+0.1*c
h is the perpendicular distance from the source to the line of mics
l is the distance from mic 1 to the perpendicular along the line of mics
θ is the angle subtended at mic 1
cos(60)=(100+l)/d2
cos(θ)=l/d1
l^2+h^2=d1^2
(100+l)^2+h^2=d2^2
(100+l)^2-l^2=d2^2-d1^2
simplify
100^2+200*l=d2^2-d1^2
recall
d2=d1+0.1*c
d1^2=d2^2-.2*c*d2+.1^2*c^2
plug it in
100^2+200*l=d2^2-d2^2+.2*c*d2-.1^2*c^2
simplify
100^2+200*l=.2*c*d2-.1^2*c^2
cos(60)=(100+l)/d2
l=d2*cos(60)-100
plug it in
200*d2*cos(60)-100^2=.2*c*d2-.1^2*c^2
solve for d2
d2=(100^2-.1^2*c^2)/(200*cos(60)-.2*c)
c=343
d2=282 m
d1=282-343*.1
d1=248 m
5)
ω is the angular speed of the light beam
ω=π rad/sec
θ=ω*t
L=6/cos(θ)
L=6/cos(π*t)
d=sqrt(L^2-36)
d=6*sqrt(1/cos^2(π*t)-1)
plug in the times for the rest of it
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