euclidean-geometrygeometrytriangles
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.
Let $AC = 2x$. Since triangle $AHC$ is half of the equliateral triangle with side $2x$ (reflect $C$ across $AB$) we see that $CH =x$.
Let $y= AB/6$ then $AH =3y$ and $AM = 2y$ so $MH = y$ and thus $${AC\over CH} = {AM\over MH}$$
so by angle bisector theorem $CM$ is angle bisector for $\angle ACH$.
Construct the equilateral triangle $AHB$. Given that $AC = BC, AH = BH$ and the shared $CH$, the triangles $AHC$ and $BHC$ are congruent. Then, $\angle BCH = \dfrac12\angle ACB = 20^\circ$.
Since $AH = BH$ and $\angle BAM = \angle HAM = 30^\circ$, the triangles $BAM$ and $HAM$ are congruent, which yields $\angle HBM = \angle BHM = \angle HBC = 10^\circ$ and $HM || CB$.
Then, the triangles $CHB$ and $BHC$ have the same altitudes $h$ with respect to the base $BC$. Since $\angle BCH = \angle CBM = 20^\circ$, we have $CH = BM = h\cot 20^\circ$.
As a result, the triangles $CHB$ and $BMC$ are congruent, which leads to,
$$\angle BMC = \angle CHB = 180^\circ - \angle CBH - \angle BCH = 180^\circ - 10^\circ - 20^\circ = 150^\circ$$
Best Answer
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.