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In my book there is this problem but I don't know if congruence or similarity can be used as means to solve it.

The problem is as follows:

A CTV broadcasting tower is resting atop a square flat base in the highest hill of Taipei. The tower is supported by five cables which are held tight to the ground by four retention screws labeled $\textrm{A}$, $\textrm{C}$, $\textrm{D}$, $\textrm{E}$, $\textrm{F}$ as indicated in the picture. The radio technician notices that the length between $\textrm{HC = 3 feet}$, meanwhile $3AB=5BC$ and $m\angle ABH= 3m \angle HBC$ From this information find the distance between $\textrm{H}$ and $\textrm{A}$.

Sketch of the problem

The alternatives given on my book are:

$\begin{array}{ll} 1.&\textrm{13 feet}\\ 2.&\textrm{11 feet}\\ 3.&\textrm{9 feet}\\ 4.&\textrm{10 feet}\\ \end{array}$

What I attempted to do is shown in the figure from below.

Sketch of the attempted solution

But other than just putting what is stated in the problem I could not reach any further in my observations. I tried hardly to imagine different ways to arrange the triangles in a manner that they could be used as congruence (from which I noticed one). But other than that I coudn't find more.

Therefore can somebody help me to find the answer using mainly simple euclidean geometry?. i.e using constructions and tying it with similarity or congruence or something like this? This is what basically gets me stuck.

Although I welcome an answer using plane trigonometry this question was supposedly been able to be solved without requiring trig or any advanced math tools, hence I hope somebody could help me with a trig free approach.

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  • $\begingroup$ I think you mean $\angle ABH = 3\angle HBC$, not $\angle ABC$ $\endgroup$
    – Quanto
    Oct 1 '19 at 0:48
  • $\begingroup$ @Quanto Yes this was a mistake from my side I corrected it in the rewording of the question. $\endgroup$ Oct 1 '19 at 20:14
  • $\begingroup$ great. then answer should work $\endgroup$
    – Quanto
    Oct 1 '19 at 20:15
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enter image description here

Draw the line BD such that $\angle DBH = \angle CBH$. Then, $DH = HC = 3$ feet. Apply the sine rule to triangles BAD and BDC,

$$\frac{\sin 2\alpha}{\sin(180-\beta)} = \frac{AH - 3}{AB}$$ $$\frac{\sin 2\alpha}{\sin\beta} = \frac{6}{BC}$$

Use the fact $\sin(180-\beta)=\sin\beta$ to get, $$\frac{AH-3}{6}=\frac{AB}{BC}=\frac 53\tag{1} $$

which yields

$$AH = 13 \>\text{feet} $$

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  • $\begingroup$ If you look closely to what I did I was not that far from your route (omiting the fact that I was not close to use trig). Can this question be solved without requiring it?. Btw i'm stuck at trying to see how did you come up with a relationship for $AB$. Can you split the equality in different lines to understand it better to know which came from where?. (sorry I'm slow with this). Perhaps do you refer to $\frac{AB}{\sin\left(180-\beta\right)}$? otherwise I still don't get it. $\endgroup$ Oct 1 '19 at 20:24
  • $\begingroup$ @ChrisSteinbeckBell - Yeah, you approach resembles that in the answer. I split the equations as you requested. $\endgroup$
    – Quanto
    Oct 1 '19 at 20:59
  • $\begingroup$ Also, the ratio Eq (1) is the result of triangle angle bisector theorem, which can be proved geometrically. Some people use it without proof. Here, I derived it with sine rule for completeness. $\endgroup$
    – Quanto
    Oct 1 '19 at 21:53
  • $\begingroup$ Thanks! In fact I believed that using solely the angle bisector teorem would suffice to solve this problem. At first looking the proof via trig confused me a bit, but once I noticed the "catch" with $\left(180-\beta\right)$ made me understood. $\endgroup$ Oct 7 '19 at 11:10

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