Two guy wires are attached to the top of a telecommunications tower and anchored to the ground on opposite sides of the tower, as shown. The length of the guy wire is 20 m more than the height of the tower. The horizontal distance from the base of the tower to where the guy wire is anchored to the ground is one-half the height of the tower. How tall is the tower, to the nearest tenth of a metre?
+26400 Two guy wires are attached to the top of a telecommunications tower and anchored to the ground on opposite sides of the tower, as shown. The length of the guy wire is 20 m more than the height of the tower. The horizontal distance from the base of the tower to where the guy wire is anchored to the ground is one-half the height of the tower. How tall is the tower, to the nearest tenth of a metre?
\(\small{ \begin{array}{rcll} \mathbf{ \text{Pythagoras:} }\\ \left( \dfrac{h}{2} \right)^2 + h^2 &=& (h+20)^2 \\ \dfrac{h^2}{4} + h^2 &=& h^2+40h+400 \\ \dfrac{h^2}{4} &=& 40h+400 \\ \dfrac{h^2}{4} -40h-400&=& 0 \qquad & | \qquad \cdot 4 \\ h^2 -160h-1600&=& 0 \\ \hline \boxed{~ \begin{array}{rcll} ax^2+bx +c &=& 0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~} \\ \hline x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \qquad a = 1 \quad b = -160 \quad c = -1600 \\ h_{1,2} &=& {160 \pm \sqrt{160^2-4\cdot(-1600)} \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160^2+40 \cdot 160 } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160\cdot (160+40) } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160\cdot 200 } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{1600\cdot 20 } \over 2} \\ h_{1,2} &=& {160 \pm 40 \sqrt{20 } \over 2} \\ h_{1,2} &=& {160 \pm 40\cdot 2\cdot \sqrt{ 5 } \over 2} \\ h_{1,2} &=& {160 \pm 80\cdot \sqrt{ 5 } \over 2} \\ h_{1,2} &=& 80 \pm 40\cdot \sqrt{ 5 } \\ h &=& 80 + 40\cdot \sqrt{ 5 } \\ h &=& 40\cdot( 2 + \sqrt{ 5 } ) \\ \mathbf{ h} & \mathbf{=} & \mathbf{169.442719100} \end{array} }\)
The tower is 169.4 m tall.
+118724 This is what I am imagining is the situation.
Now use Pythagorean theorum.
\((t+20)^2=t^2+(t/2)^2\\ t^2+40t+400=(4t^2+t^2)/4\\ 4t^2+160t+1600=5t^2\\ 0=t^2-160t-1600\\\)
\(x = {160 \pm \sqrt{160^2+4*1600} \over 2}\\ x = 169.4427\)
The height is 169.4 m
+26400 Two guy wires are attached to the top of a telecommunications tower and anchored to the ground on opposite sides of the tower, as shown. The length of the guy wire is 20 m more than the height of the tower. The horizontal distance from the base of the tower to where the guy wire is anchored to the ground is one-half the height of the tower. How tall is the tower, to the nearest tenth of a metre?
\(\small{ \begin{array}{rcll} \mathbf{ \text{Pythagoras:} }\\ \left( \dfrac{h}{2} \right)^2 + h^2 &=& (h+20)^2 \\ \dfrac{h^2}{4} + h^2 &=& h^2+40h+400 \\ \dfrac{h^2}{4} &=& 40h+400 \\ \dfrac{h^2}{4} -40h-400&=& 0 \qquad & | \qquad \cdot 4 \\ h^2 -160h-1600&=& 0 \\ \hline \boxed{~ \begin{array}{rcll} ax^2+bx +c &=& 0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~} \\ \hline x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \qquad a = 1 \quad b = -160 \quad c = -1600 \\ h_{1,2} &=& {160 \pm \sqrt{160^2-4\cdot(-1600)} \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160^2+40 \cdot 160 } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160\cdot (160+40) } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160\cdot 200 } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{1600\cdot 20 } \over 2} \\ h_{1,2} &=& {160 \pm 40 \sqrt{20 } \over 2} \\ h_{1,2} &=& {160 \pm 40\cdot 2\cdot \sqrt{ 5 } \over 2} \\ h_{1,2} &=& {160 \pm 80\cdot \sqrt{ 5 } \over 2} \\ h_{1,2} &=& 80 \pm 40\cdot \sqrt{ 5 } \\ h &=& 80 + 40\cdot \sqrt{ 5 } \\ h &=& 40\cdot( 2 + \sqrt{ 5 } ) \\ \mathbf{ h} & \mathbf{=} & \mathbf{169.442719100} \end{array} }\)
The tower is 169.4 m tall.
