Basic College Mathematics (10th Edition)

(a) The 2-meter stick is $52.5$ meters from the tower. (b) The 1-meter stick must be placed $55.25$ meters from the tower.
(a) The distance of the 2-m stick from the tower can be found by subtracting $3.5$ meters from $56$ meters: distance = $56 - 3.5 = 52.5$ meters (b) The 1-meter stick has to be placed somewhere between the 2-meter stick and the rightmost point (or vertex) of the biggest triangle. This will form the smallest right triangle that is similar to the existing two right triangles in the given illustration. Let $x$ = distance of the 1-meter stick from the biggest triangle's vertex at the right Using similar triangles, we'd have: $$\dfrac{\text{height of the smallest right triangle}}{\text{height of the medium right triangle}} = \dfrac{\text{base of the smallest right triangle}}{\text{base of the medium right triangle}} \\\dfrac{1}{2}=\dfrac{x}{3.5} \\2(x) = 1(3.5) \\2x=3.5 \\x=\frac{3.5}{2} \\x=1.75$$ Thus, the distance of the 1-meter stick from the tower is: distance $= 56 - 1.75 = 55.25$ meters