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An isosceles triangle $\triangle ABC$ is given with $\angle ACB=30^\circ$ and leg $BC=16$ $cm$. Find the perimeter of $\triangle ABC$.

We have two cases, right? When 1) $AC=BC=16$ and 2) $AB=BC=16$.

For the first case: let $CH$ be the altitude through $C$. Since the triangle is isosceles, $CH$ is also the angle bisector and $\measuredangle ACH=\measuredangle BCH=15^\circ$. How to approach the problem further? I have not studied trigonometry.

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The first case.

Let $BK$ be an altitude of $\Delta ABC$.

Thus, $$BK=8,$$ $$CK=\sqrt{BC^2-BK^2}=\sqrt{16^2-8^2}=8\sqrt3$$ and $$AB=\sqrt{AK^2+BK^2}=\sqrt{(16-8\sqrt3)^2+8^2}=$$ $$=\sqrt{8^2(2-\sqrt3)^2+8^2}=8\sqrt{(2-\sqrt3)^2+1}=16\sqrt{2-\sqrt3},$$ which gives the answer: $32+16\sqrt{2-\sqrt3}.$

We can solve the second problem by the similar way.

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  • $\begingroup$ Thank you for the response! I didn't see that if we construct an altitude through $A$ or $B$ we can find it. $\endgroup$ – Katherine May 13 '20 at 17:50
  • $\begingroup$ Can I ask you why $\sqrt{(16-8\sqrt3)^2+8^2}=8\sqrt{(2-\sqrt3)^2+1}$? $\endgroup$ – Katherine May 13 '20 at 17:51
  • $\begingroup$ How did you see this that fast? $\endgroup$ – Katherine May 13 '20 at 17:53
  • $\begingroup$ For the second case of the problem, I found $P_{\triangle ABC}=32+16\sqrt{3}$, right? $\endgroup$ – Katherine May 13 '20 at 17:59
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    $\begingroup$ @LYI I added something,See now. $\endgroup$ – Michael Rozenberg May 13 '20 at 19:16
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The only fact that you need is that the right triangle with sides $1$,$\sqrt{3}$ and hypothenuse $2$ has angles $30$, $60$.

Have this picture in mind: The right triangle $MNP$ with $\angle NMP=30$, $\angle MPN=60$ and $\angle MNP=90$ with $NP=1$, $MN=\sqrt{3}$ and $MP=2$.

Now, prolong the line $\vec{NM}$ until you reach the point $Q$ such that $MQ=MP=2$. With this construction $QMP$ is an isosceles triangle, so $\angle MQP=\angle MPQ$. Since $\angle MQP+\angle MPQ=30$, we have that in fact $\angle MQP=\angle MPQ=15$. Now, note that we have a right triangle $QPN$, with $\angle NQP=15$, and their sides are $PN=1$, $QN=2+\sqrt{3}$. construction of right triangle with one of its angles being 15 By pythagoras, you have the right triangle $PNQ$ with $PN=1$, $NQ=2+\sqrt{3}$ and hypothenuse $2\sqrt{2+\sqrt{3}}$.

Going back to your first problem, you have a right triangle with angle $15$ and hypothenuse $BC=16$, By similarity of triangles you can get $HB$ (namely, $\frac{HB}{BC}=\frac{1}{2\sqrt{2+\sqrt{3}}}$) Can you get it from here?

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  • $\begingroup$ Even more, "the fact" that I mentioned at the beginning can be easily proved by tracing the altitude from an equilateral triangle of length side $2$. $\endgroup$ – Julian Mejia May 13 '20 at 16:50
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No trigs.

Case 1. $|AB|=|BC|=16$.

enter image description here

This case is simple, $\triangle BCE$ is equilateral, $|BD|,\ |CD|$ and $|AC|$ can be easily found.

Case 2. $|AC|=|BC|=16$.

This case is just a couple of steps longer.

Extend $BD$ such that $|DE|=|BD|$. Then $\triangle BCE$ is equilateral.

enter image description here

\begin{align} \triangle BCD:\quad |BD|&=\tfrac12\,|BC|=8 ,\\ |CD|&=\sqrt{|BC|^2-|BD|^2} =8\,\sqrt3 ,\\ |AD|&=|AC|-|CD|=16-8\,\sqrt3 ,\\ |AB|&=\sqrt{|AD|^2+|BD|^2} =8\sqrt2(\sqrt3-1) . \end{align}

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  • $\begingroup$ Really interesting! Thank you for the response! I appreciate it. $\endgroup$ – Katherine May 13 '20 at 19:14
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Can you use the cosine formula?

$$ \cos(A)=\frac{b^2+c^2-a^2}{2bc}$$

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  • $\begingroup$ I have not studied trigonometry! $\endgroup$ – Katherine May 13 '20 at 16:28
  • $\begingroup$ thats the easiest approach according to me $\endgroup$ – Dudeness May 13 '20 at 16:29
  • $\begingroup$ This does not change the fact that I am not familiar with that concept. $\endgroup$ – Katherine May 13 '20 at 16:30
  • $\begingroup$ Mind your language, please! I HAVE NOT STUDIED trigonometry. This includes trig functions. $\endgroup$ – Katherine May 13 '20 at 16:33
  • $\begingroup$ and mind your language what? $\endgroup$ – Dudeness May 13 '20 at 16:34

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