Did you make your sketch?
I labeled exit 81 West as A and exit 74 East as B
so AB = 7 km
label the position of the balloon as P and the point below as Q
PQ = .875 km
In triangle PAQ .875/AQ = tan 11
AQ = .875/tan11 = ....
so QB = 7 - AQ
tanØ = 875/QB = ....
then find Ø
in triangle PBQ
Mike's hot-air balloon is 875.0 m directly above a highway. When he is
looking west, the angle of depression to Exit 81 is 11 °. The exit
numbers on this highway represent the number of kilometers left
before the highway ends. What is the angle of depression, to the
nearest degree, to Exit 74 in the east?
2 answers
distance between the exits is ... d1 = 81 km - 74 km
road distance from balloon to x-81 ... d2 = 875.0 m / tan(11º)
road distance to x-74 ... d3 = d1 - d2
angle of depression to x-74 ... tan(Θ) = 875.0 m / d3
road distance from balloon to x-81 ... d2 = 875.0 m / tan(11º)
road distance to x-74 ... d3 = d1 - d2
angle of depression to x-74 ... tan(Θ) = 875.0 m / d3