base of the building is 45 meters. How many meters high is the ledge?
2006-12-12 11:03:02 · 4 answers · asked by Min 1 in Science & Mathematics ➔ Mathematics
Remember SOHCAHTOA! (Sine equals Opposite over Hypotenuse, Cosine equals Adjacent over Hypotenuse, Tangent equals Opposite over Adjacent)
Here we have the adjacent (A) and we are looking for the opposite (O), so we need the Tangent:
tan(x) = Opposite / Adjacent
tan(27º) = Opposite / 45 m
Opposite = tan(27º) * 45 m
Opposite = 22.9 meters
Therefore, the ledge is 22.9 m high.
2006-12-12 11:05:48 · answer #1 · answered by sep_n 3 · 0⤊ 0⤋
Make a triangle to help you.
Label the base 45 m, the hypotneuse y, and the other side as x.
The angle from the base to the ledge is 27o, so set up an equation using tan.
x/ 45= tan 27o
x= 45 tan 27o= 22.93 m high.
2006-12-12 11:08:34 · answer #2 · answered by mini_roller 3 · 0⤊ 0⤋
Let the height of the building = h.
Then:
tan 27° = h/45
h = 45 tan 27° = 22.93 meters high
2006-12-12 11:11:33 · answer #3 · answered by Northstar 7 · 0⤊ 0⤋
the height h=45 tan 27*
2006-12-12 11:04:55 · answer #4 · answered by raj 7 · 0⤊ 0⤋