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.
$\endgroup$4 Answers
$\begingroup$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.
$\endgroup$ 9 $\begingroup$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 $PN=1$, $QN=2+\sqrt{3}$. 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?
$\endgroup$ 1 $\begingroup$No trigs.
Case 1. $|AB|=|BC|=16$.
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.
\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}
$\endgroup$ 1 $\begingroup$Can you use the cosine formula?
$$ \cos(A)=\frac{b^2+c^2-a^2}{2bc}$$
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