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August 30, 2014

August 30, 2014

Posted by **Anonymous** on Wednesday, July 15, 2009 at 12:57pm.

(i) the height of the vertical ladder

(ii) the distance between the building and the ladder

- Math -
**MathMate**, Wednesday, July 15, 2009 at 7:58pm*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 fire-truck.

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}+(d-r)^{2})

=12 + sqrt(144*2 + (13.856-12)^{2})

=12 + sqrt(288+ 3.446)

= 12+17.07

= 29.07 m.

- Math -
**MathMate**, Wednesday, July 15, 2009 at 8:07pmA sketch is available at the following link:

http://i263.photobucket.com/albums/ii157/mathmate/Trigo.jpg

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