On a hill, inclined at 15.4 degrees with the horizontal, stands a vertical tower. At a point Q, 61.5 inches down the hill from the foot of the tower, the angle of elevation of the top of the tower is 42.6 degrees. How tall is the tower?
2007-10-26 01:37:24 · 2 answers · asked by delixir_21 1 in Science & Mathematics ➔ Mathematics
For this problem, you have to solve two right triangles trig functions.
base of triangle of tower+elev: 61.5*cos(15.4)= 59.29
length of triangle of tower+elev: 59.29*tan(42.6)=54.52
length of elev: 61.5*sin(15.4)=16.33
height of tower = 54.52 - 16.33 = 38.19 inches
2007-10-26 02:24:01 · answer #1 · answered by Parallels 2 · 0⤊ 0⤋
The triangle formed by the point Q,the top of the tower and its foot has angles 42.6, 105.4 and 32 º
so using the sine theorem
h/sin42.6 = 61.5/sin 32
so
h=78.56 inches
2007-10-26 14:40:50 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋