Cate is sailing her boat off the coast, which
runs straight north and south. Her GPS
confirms that she is 8 km from Haytown
and 10 km from Beeville, two towns on the
coast. The towns are separated by an angle
of 80°, as seen from the boat. A helicopter
is hovering at an altitude of 1000 m
halfway between Haytown and Beeville.
b) Determine the angle of elevation of the
helicopter, as seen from the sailboat, to
the nearest tenth of a degree.
2 answers
The length between the two towns is 11.67km if that helps
First we find the distance between (H)aytown and (B)eebille
AB^2 = 8^2 + 10^2-2(8)(10)cos80°
AB = 11.67117
let the midpoint be M
then BM = 5.835586....
Let S be the position of the ship,
again, by the cosine law:
let's find angle B
8^2 = 10^2 + 11.67117..^2 - 2(10)(11.67117)cosB
cosB = .73778..
B = 42.457°
Once more by the cosine law in triangle SBM
SM^2 = 10^2 + 5.835586..^2 - 2(10)(5.835586..)cos42.457
SM = .... , you do it
finally we can find our angle
tan(angle) = 1000/SM = ...
take tan^-1 to find the angle of elevation
AB^2 = 8^2 + 10^2-2(8)(10)cos80°
AB = 11.67117
let the midpoint be M
then BM = 5.835586....
Let S be the position of the ship,
again, by the cosine law:
let's find angle B
8^2 = 10^2 + 11.67117..^2 - 2(10)(11.67117)cosB
cosB = .73778..
B = 42.457°
Once more by the cosine law in triangle SBM
SM^2 = 10^2 + 5.835586..^2 - 2(10)(5.835586..)cos42.457
SM = .... , you do it
finally we can find our angle
tan(angle) = 1000/SM = ...
take tan^-1 to find the angle of elevation
7 answers