After the spaceship's arrival, a couple of bored New Yorkers decided to figure out how high up it was hovering. One stood at the north end of Manhattan, and measured the angle of elevation to the center of the ship to be 6.1∘6.1∘; the other stood at the south end of the island, and measured the angle of elevation to the center of the ship to be 8.1∘8.1∘. If the two people were 13.413.4 miles apart, how high up was the spaceship, in miles? (Nothing here says that the center of the ship is halfway between the two people!)
im getting either 7.6968 an 5.8047, and its still wrong
+129839 We can first use this..... [ We're using 13.413 mi......??? ]
tan (6.1) = h/x → h = x*tan(6.1)
And then we have
tan (8.1) = h/ [13.413 - x] = [ x*tan(6.1)] / [13.413 - x] ....so we have
tan (8.1) = [ x*tan(6.1)] / [13.413 - x]
[13.413 - x] tan (8.1) = x*tan(6.1)
13.413 tan (8.1) - x* tan (8.1) = x* tan (6.1)
13.413 tan (8.1) = x* tan (8.1) + x* tan (6.1)
13.413 tan (8.1) = x * [ tan (8.1) + tan (6.1)]
x = 13.413 tan (8.1) / [ tan (8.1) + tan (6.1)] ≈ 7.66 mi
And the height is x*tan(6.1) = 7.66 * tan(6.1) ≈ .8187 mi
Here's a pic :
+37084 Another method (see diagram)
Law of sines : 13.413/sin 165.8 = x/sin6.1 x = 5.8103 m
then sin 8.1 = H/5.8103 H = .81867 m
