How can I find the length of a segment formed by the shadow of a ladder?

Jun 2017
399
6
Lima, Peru
The problem is as follows:

Roger is an interior designer and one morning he sees that his ladder is making a shadow with respect to the floor in the backyard as indicated in the diagram from below. He's curious and found that the angles formed between the corners are the following: $\frac{\angle CAB}{2}=\frac{\angle ADC}{2}=\frac{\angle ABC}{3}=\frac{\angle ACD}{8}$. He also notices that $AB=CD$ and $AC=8$. Find the length of the segment AD which is seen by Roger.



$\begin{array}{ll}
1.&16\,cm\\
2.&20\,cm\\
3.&15\,cm\\
4.&21\,cm\\
\end{array}$

I'm not sure exactly what sort of theorem or identity should be used here. I attempted different way to tackle this problem. Can someone help me?.

I'm slow at seeing things in space, so an answer which would help me the most is one which can be detailed the most as possible. This problem belongs to a section of euclidean geometry. So I appreciate that an answer could follow an euclidean geometry approach and a trigonometrical one as well so I can compare both methods and see which one is quicker. So can someone help me?.
 

skipjack

Forum Staff
Dec 2006
21,481
2,470
For convenience, I'll use $D$ to denote $\angle ADC$.
From the given information, $\angle ACD = 4D$, $\angle ABC = 3D/2$ and $\angle CAB = D$.
Also, $\sin(\angle CAD) = \sin(180^\circ - \angle ACD - D) = \sin(5D)$.

As $AD/\sin(\angle ACD) = CD/\sin(\angle CAD) = AC/\sin(\angle ADC) = 8\text{cm}/\sin(D)$, $CD = 8\text{cm}\sin(5D)/\sin(D)$ and $AD = 8\text{cm}\sin(4D)/\sin(D)$.

As $AB/\sin(180^\circ - \angle CAB - \angle ABC) = AC/\sin(\angle ABC) = 8\text{cm}/\sin(3D/2)$,
$AB = 8\text{cm}\sin(5D/2)/\sin(3D/2)$.

You are given that $AB = CD$, so $\sin(5D)/\sin(D) = \sin(5D/2)/\sin(3D/2)$.
As $\sin(5D) = 2\sin(5D/2)\cos(5D/2)$, $2\cos(5D/2)\sin(3D/2) = \sin(D)$.
By a well-known product-to-sum identity, $2\cos(5D/2)\sin(3D/2) = \sin(4D) - \sin(D)$,
so $\sin(4D) = 2\sin(D)$.
Hence $AD = 8\text{cm}\sin(4D)/\sin(D) = 16\text{cm}$.

It would seem that Roger was using a toy ladder.
 
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Jun 2017
399
6
Lima, Peru
For convenience, I'll use DD to denote ∠ADC∠ADC.
From the given information....

It would seem that Roger was using a toy ladder.
I don't know whether it is me, but why does it seem that the approach seems too complicated? Doesn't there exist a trig-free approach to this? Whoever did this question didn't take into consideration the meaning of the metric system. I believe it was meters not centimeters.
How can it be solved relying on Euclidean geometry?

I really tried to follow the sequence but the formatting doesn't really help much. Can you change it into fractions avoiding slashes? Perhaps can you help me with that? (Note: Sorry I had to trim your original message. For some reason, I got a message from the forum which forbade me from inserting your full answer telling me that it was too long).
 

skipjack

Forum Staff
Dec 2006
21,481
2,470
There's no need to quote any of my post. I hope the following is easier on the eye.

As \(\displaystyle \frac{AD}{\sin(\angle ACD)} = \frac{CD}{\sin(\angle CAD)} = \frac{AC}{\sin(\angle ADC)} = \frac{8\text{cm}}{\sin(D)}\),

\(\displaystyle CD = 8\text{cm}\frac{\sin(5D)}{\sin(D)}\) and \(\displaystyle AD = 8\text{cm}\frac{\sin(4D)}{\sin(D)}\).

As \(\displaystyle \frac{AB}{\sin(180^\circ - \angle CAB - \angle ABC)} = \frac{AC}{\sin(\angle ABC)} = \frac{8\text{cm}}{\sin(3D/2)}\),

\(\displaystyle AB = 8\text{cm}\frac{\sin(5D/2)}{\sin(3D/2)}\).

You are given that $AB = CD$, so \(\displaystyle \frac{\sin(5D)}{\sin(D)} = \frac{\sin(5D/2)}{\sin(3D/2)}\).
 
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