Ladder on the Stairs

By Pythagoras, the height of the taller right-triangle shown is $\sqrt{(150+d)^2-45^2}$ . Then by similar triangles we get an equation for $d$ :

$\cfrac{150+d}{\sqrt{(150+d)^2-45^2}}=\cfrac{150-d}{30}$ , which can be solved numerically for $d\approx 119.573$ .

Continuing the lines per attached sketch, we can notice there is perfect symmetry to vertical axis CI (virtual symmetrical stairs and symmetrical line DE to AB drawn). Line AB (slope: 30 vertical, 45 horizontal, that is 10 vert,15 horiz) passes thru mid risers and mid treads of steps (risers 10cm, treads 15 cm). Both A and B are in the middle of a tread. Using the symmetry, we see that both AB and vertical axis CI pass thru mid tread of second step. Horizontal distance AI is exactly 2 treads = 30cm, base AD of isosceles triangle ACD is 60 cm. So:

$h^2 + 30^2 = 150^2 \text{( right triangle ACI or DCI)}\\ h^2 = 21,600\\ h = 146.97 \text{( keeping 2 decimals)}\\ \text{also} \cos a = \frac{146.97}{150} = 0.9798\\ x = h-30 = 116.97\\ BC = \frac{x}{cos a} = \frac{116.97}{0.9798} = \underline{119.382} \text{ (keeping 3 decimals)}$

In my solution, I have used a trigonometric approach.

I have labelled the three vertices of interest as A, B and C, as shown in the above diagram. I have also connected the side AB.

Let $\phi$ be the angle that AC makes with the horizontal. Note that because of the imposed condition, $\phi$ is also the angle that BC makes with the horizontal.

With this, and taking A to be at the origin, it follows that the coordinates of C and B are given by

$C = ( 150 \cos \phi , 150 \sin \phi )$ ,

$B = ( 45, 30 )$

The condition on the legs of the ladder implies that the slope of AC is the negative of the slope of BC.

$\text{Slope of } AC = tan \phi = \dfrac{\sin \phi }{\cos \phi }$

$\text{Slope of } BC = \dfrac{(150 \sin \phi - 30)}{(150 \cos \phi - 45) }$

Therefore,

$- \dfrac{\sin \phi}{\cos \phi} = \dfrac{(150 \sin \phi - 30)}{(150 \cos \phi - 45) }$

Cross multiplying, we obtain,

$\sin \phi (45 - 150 \cos \phi) = \cos \phi (150 \sin \phi - 30 )$

Expanding and re-arranging,

$300 \sin \phi \cos \phi - 30 \cos \phi - 45 \sin \phi = 0$

Using the double angle sine formula, the above becomes,

$150 \sin 2 \phi - 30 \cos \phi - 45 \sin \phi = 0$

This equation can be solved using Newton's method for finding function roots. It goes as follows

Let $f( \phi ) = 150 \sin 2 \phi - 30 \cos \phi - 45 \sin \phi$ be the function we want to find the root of.

First calculate the derivative, $f'( \phi) = 300 \cos 2 \phi + 30 \sin \phi - 45 \cos \phi$

Secondly start with an initial guess $\phi_0$ , and calculate successive approximations of the root using the following formula (Newton's Iteration)

$\phi_{n+1} = \phi_n - \dfrac{ f(\phi_n) }{f'(\phi_n) }$

I started with $\phi_0 = \dfrac{ \pi }{3}$

The method converged in 5 iterations to $\phi = 1.403080542$ rad.

Now the length of BC can be calculated from the distance formula , as:

$BC = \sqrt{ (150 \cos \phi - 45 )^2 + (150 \sin \phi - 30 )^2 } = \boxed{119.573}$

Let assign $h$ , $a$ and $\theta$ as in the figure. Then $h = 150 \cos \theta$ and $a = 150 \sin \theta$ . We note that

$\begin{aligned} \tan \theta & = \frac ah = \frac {45-a}{h-30} \\ \frac {\sin \theta}{\cos \theta} & = \frac {45 - 150\sin \theta}{150\cos \theta - 30} \\ 150\sin \theta \cos \theta - 30 \sin \theta & = 45 \cos \theta - 150\sin \theta \cos \theta \\ 300\sin \theta \cos \theta - 30 \sin \theta - 45 \cos \theta & = 0 \\ 20\sin \theta \cos \theta - 2 \sin \theta - 3 \cos \theta & = 0 \end{aligned}$

Solving the equation numerically, we get $\theta \approx 0.1677158 \text{ rad}$ . Then, we have

$x = \begin{cases} \dfrac {45-a}{\sin \theta} = \dfrac {45-150\sin \theta}{\sin \theta} \\ \dfrac {h-30}{\cos \theta} = \dfrac {150\cos \theta-30}{\cos \theta} \end{cases} \approx \boxed{119.573}$

Using the law of cosines, one can compute the angles at the bottom of the triangle, $150^2 = 150^2 + 45^2 - 2 \cdot 150 \cdot 45 \cdot cos(A)$. A = 81.37 deg. Finally one can calculate the length of the size in question using the following equation (definition of sine) $sin(81.37)=\frac{30}{150-k}$. k=119.573

Make a right triangle out of the longer leg by creating a vertical line down from the apex of the ladder. We can calculate the height of that line this way:

150^2 = 22.5^2 + h^2

h= 148.3

Now we can create a right triangle with the shorter leg in the same fashion and the height of that will be 148.3 - 30 = 118.3.

So the two legs of a right triangle formed with the short leg as the hypotenuse are 22.5 and 118.3. So now the square root of (22.5^2 + 118.3^2) is the length of the short leg = 120.4 cm.

Consider my diagram. By pythagorean theorem, $h=\sqrt{150^2-22.5^2}=\sqrt{21993.75}$

$\sin \theta = \dfrac{h}{150}=\dfrac{h-30}{x}$ $\implies$ $x=\dfrac{150(h-30)}{h}=\dfrac{150(\sqrt{21993.75}-30)}{\sqrt{21993.75}} \approx \boxed{119.657}$

Where the "lines" touch they are close to the middle. So you could get rid of the diagonal element and then do the equation. There are 3 10-inch steps and 3×10=30 so subtract 30 from 150 and you get 120in.

Draw a perpendicular and then apply Pythagoras to both left and right triangles

Suppose if we extend the shorter leg up to the same level as that of bigger one Then vertical height adds on to 30units more.. and now we if could imagine up the scenario and take increased vertical height same as ladders extended length, it nearly Coincides (though it's not equal but for simplification) then as congruency angles would be same and so the opposite lengths. So, x+30=150; x=150-30=120.

The shorter length will nearly be equal to 120 units.