Math
posted by Anonymous .
A building 24 m tall is viewed from the top and bottom of a vertical ladder.The building forms an angle of 45 degrees with the top of the ladder and an angle of 60 degrees with the bottom of the ladder.Find
(i) the height of the vertical ladder
(ii) the distance between the building and the ladder

The top of the building forms an angle of 45 degrees with the top of the ladder...
I believe the question means the top of the building makes 45 and 60 degrees with the horizontal, and not the building itself (i.e. from bottom to top).
In this case, the distance from the building is
d=24 cot(60)=8sqrt(3).
The height of the ladder is
24m  d.tan(45)
= 24  8sqrt(3)
= 10.144 m.
If the angles are subtented by the top and bottom of the building, the feasible solution of the height of the ladder is higher than the building, as if it is a ladder from a firetruck.
In this case, we calculate the centre of a circle which passes through the top and bottom of the building, and subtends an angle of 90 deg. It is a point at 12 m from the face of the building and 12 m. high above ground. The radius of the circle is r=12sqrt(2) m.
From here we find the intersection with the ladder at d=8sqrt(3) from the building. There is one intersection above ground at a height of
h = 12m + sqrt(r^{2}+(dr)^{2})
=12 + sqrt(144*2 + (13.85612)^{2})
=12 + sqrt(288+ 3.446)
= 12+17.07
= 29.07 m. 
A sketch is available at the following link:
http://i263.photobucket.com/albums/ii157/mathmate/Trigo.jpg