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The problem is as follows:

The figure from below shows a certain terrain has the shape of a triangle $\triangle ABC$ which is obtuse on $B$. Assuming the line connecting $AM$ is a bisector whose angle is $\beta$ and $\angle ACB=20^{\circ}$ and $\angle ABC= 120^{\circ}$ and $\angle CMA=90^{\circ}$ and $CM=8\sqrt{3}\,m$. Find the cost of placing a fence on $CM$ if the owner of the terrain consulted with the hardware store and the price is $\$3.0$ per meter.

Sketch of the problem

The alternatives in my book are as follows:

$\begin{array}{ll} 1.&\$\,21.00\\ 2.&\$\,24.00\\ 3.&\$\,48.00\\ 4.&\$\,32.00\\ \end{array}$

In my attempt to solve this problem the only thing which I was able to spot is that the bisector angle is $\beta= 20^{\circ}$.

This comes because of the sum of the interior angles in the triangle add up to $180^{\circ}$.

Hence on $\triangle ABC$:

$2\beta+120+20=180$

$\beta= 20^{\circ}$

Then I could also spot is that the small right triangle has the interior angles of 40^{\circ} and 50^\circ because by interior and exterior angle on the smaller triangle. But other than that I cannot spot something else.

Can someone help me?. I think in this problem is required congruence or similarity but I don't know how to use those here.

Please an answer must include a drawing Since it is not easy for me to spot the relationships and construction that would help me better to understand what's going on.

1 Answers1

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Extend the line BC to the left until it comes "under" point A. Then drop a line from point A to the extended line BC so that the line is perpendicular to BC. Denote the point of intersection as E.

Then $\triangle$ AEC is a 20-70-90 right triangle.

Since triangles AEC and ACM have similar angles, and since they share the long side of AC,

$$ \text{AE is also} ~8\sqrt{3}.$$

Since $\triangle$ AEB is a 30-60-90 right triangle,

$$ \text{AB is} ~16.$$

Since the cost is 3 dollars per meter, the cost of the fence is 48 dollars.

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