my answer is:
sin 55 = 20/y
y=20/sin55
y=24.42
55-28=27
sin27/x=sin28/24.42
24.42sin27/sin28=23.61
is my answer correct?
a passenger in an airplane at an altitude of a = 20 kilometers sees two towns directly to the east of the plane. The angle of depression to the towns are 55˚ and 28˚. How far apart are the towns?
2 answers
No idea what x and y are, you did not define them.
Since the altitude would be be measured perpendicular to the ground
label the position of the plane P and the bottom of the altitude as Q
Label the towns with angles of depression of 55˚ and 28 as A and B respectively.
tan 55 = 20/QA
QA = 20/tan55 = 14.004 km
tan28 = 20/QB
QB = 20/tan28° = 37.615 km
distance between towns = QB - QA = ....
It looks like you first found the hypotenuse of the smaller right-angled triangle
then used the sine law to find the distance between the towns
However, the way you presented the solution would not be acceptable.
1. no definitions of x and y
2. 55-28=27 <---- I had to guess what that calculation was supposed to be
3. from: sin27/x=sin28/24.42
to : 24.42sin27/sin28=23.61 you lost your variable
Since the altitude would be be measured perpendicular to the ground
label the position of the plane P and the bottom of the altitude as Q
Label the towns with angles of depression of 55˚ and 28 as A and B respectively.
tan 55 = 20/QA
QA = 20/tan55 = 14.004 km
tan28 = 20/QB
QB = 20/tan28° = 37.615 km
distance between towns = QB - QA = ....
It looks like you first found the hypotenuse of the smaller right-angled triangle
then used the sine law to find the distance between the towns
However, the way you presented the solution would not be acceptable.
1. no definitions of x and y
2. 55-28=27 <---- I had to guess what that calculation was supposed to be
3. from: sin27/x=sin28/24.42
to : 24.42sin27/sin28=23.61 you lost your variable
2 answers