A photographer points a camera at a window in a nearby building forming an angle of 42 with the camera platform. If the camera is 52 m from the building, how high above the platform is the window, to the nearest hundredth of a meter?
Thanks.
2007-11-23 00:19:21 · 7 answers · asked by labelapark 6 in Science & Mathematics ➔ Mathematics
tan42=x/52
therefore:
x=(tan42)(52)
2007-11-23 00:24:44 · answer #1 · answered by rAvEn JoHn 2 · 2⤊ 0⤋
Treat this as a Right Angled Triangle where you are aware of;
Horizontal distance = 52 metres {AB}
Angle from Camera to building = 42 Degrees {Angle ABC}
Height = AC
Tan 42 = [AC / AB]
So AC = Tan42 x AB
Use Tables / Calculator Tan 42 = 0.9004
AC = 52 x 0.9004 = 46.82 metres
2007-11-23 08:50:05 · answer #2 · answered by Rod Mac 5 · 0⤊ 0⤋
Let height of building = h m
tan 42° = h / 52
h = 52 tan 42°
h = 46.82 m
2007-11-28 10:26:22 · answer #3 · answered by Como 7 · 0⤊ 1⤋
Let the height is h.
tan42 = h / 52
h = (tan 42) * 52
Use your calculator for this one.
h = 46.82 m
2007-11-23 08:30:34 · answer #4 · answered by ? 6 · 0⤊ 0⤋
do an experiment to find out
2007-11-23 08:23:06 · answer #5 · answered by Anonymous · 0⤊ 5⤋
hey dude.. can you wait for my answer.... i will answer it for you.. i will be right back
2007-11-23 08:25:38 · answer #6 · answered by Anonymous · 1⤊ 4⤋
ask your teacher :)
2007-11-23 08:21:58 · answer #7 · answered by Anonymous · 0⤊ 5⤋