K
khorne
Guest
Hey, I've been working at a few problem in the fitzpatrick book, but the only way I can see to solve them is by using areas, and I'm certain there must be an easier way:
Problem:
Two buildings of equal height are 40m apart and at a point on the horizontal line joining their feet, the angles of elevation to the tops of the buildings are 47 and 28. Show that the height h of the buildings is given by
h = [40 tan47 tan28]/[tan47 + tan28]
Working:
Draw a diagram
Area of rectangle (with towers as sides) = 40h
Area of triangle (base 40, height h) = 20h, therefore area taken up by the triangles formed with the towers as sides is 20h
let the foot of one tower to the point on the horizontal = x and the other = y
Therefore y = hsin62/sin28 and x = hsin43/sin47
So the area of those is equal to:
A(y) = h^2sin62/2sin28
A(x) = h^2sin43/2sin47
Now adding them, and equating to 20h, and then simplify etc, gives the right identity...But I was wondering if there was another way to do it? Any ideas?
Problem:
Two buildings of equal height are 40m apart and at a point on the horizontal line joining their feet, the angles of elevation to the tops of the buildings are 47 and 28. Show that the height h of the buildings is given by
h = [40 tan47 tan28]/[tan47 + tan28]
Working:
Draw a diagram
Area of rectangle (with towers as sides) = 40h
Area of triangle (base 40, height h) = 20h, therefore area taken up by the triangles formed with the towers as sides is 20h
let the foot of one tower to the point on the horizontal = x and the other = y
Therefore y = hsin62/sin28 and x = hsin43/sin47
So the area of those is equal to:
A(y) = h^2sin62/2sin28
A(x) = h^2sin43/2sin47
Now adding them, and equating to 20h, and then simplify etc, gives the right identity...But I was wondering if there was another way to do it? Any ideas?
