MHB Possibilities for the side and angle of the triangle

[mathmari]

Hey!

We have the triangle ABC with $c=18$, $\alpha=\frac{\pi}{6}$ and the median through $C$ has the length $5$. I want to determine all the possibilities for $a$ and $\gamma$.

I have done the following:

From the median we have the following:
$$m_c=\sqrt{\frac{2a^2+2b^2-c^2}{4}}\Rightarrow m_c^2=\frac{2a^2+2b^2-c^2}{4} \Rightarrow 4m_c^2=2a^2+2b^2-c^2 \\ \Rightarrow 4\cdot 25=2a^2+2b^2-18^2 \Rightarrow 2a^2+2b^2=100+324 \Rightarrow 2a^2+2b^2=424 \\ \Rightarrow a^2+b^2=212 \Rightarrow b^2=212-a^2$$

From the cosine law we have that \begin{align*}&a^2=b^2+c^2-2\cdot b\cdot c\cdot \cos \alpha \\ & \Rightarrow a^2=212-a^2+c^2-2\sqrt{212-a^2}\cdot c\cdot \frac{\sqrt{3}}{2} \\ & \Rightarrow 2a^2=212+324-18\sqrt{3}\sqrt{212-a^2} \\ & \Rightarrow 2a^2=536-18\sqrt{3}\sqrt{212-a^2} \\ & \Rightarrow a^2=268-9\sqrt{3}\sqrt{212-a^2} \\ & \Rightarrow 9\sqrt{3}\sqrt{212-a^2}=268-a^2 \\ & \Rightarrow \left (9\sqrt{3}\sqrt{212-a^2}\right )^2=\left (268-a^2\right )^2 \\ & \Rightarrow 81\cdot 3\cdot \left (212-a^2\right )=71824-536a^2+a^4 \\ & \Rightarrow 243\cdot \left (212-a^2\right )=71824-536a^2+a^4 \\ & \Rightarrow 51516-243a^2=71824-536a^2+a^4 \\ & \Rightarrow a^4-293a^2+20308=0 \\ & \Rightarrow \left (a^2\right )^2-293\left (a^2\right )+20308=0 \\ & \Rightarrow a^2=\frac{-(-293)\pm \sqrt{(-293)^2-4\cdot 1\cdot 20308}}{2\cdot 1} \\ & \Rightarrow a^2=\frac{293\pm \sqrt{85849 -81232 }}{2} \\ & \Rightarrow a^2=\frac{293\pm \sqrt{4617 }}{2} \\ & \Rightarrow a^2=\frac{293\pm 9\sqrt{57}}{2} \\ & \Rightarrow a=\frac{293\pm 9\sqrt{57}}{2}\end{align*}

From the sine law we have that \begin{align*}&\frac{\sin\alpha}{a}=\frac{\sin \gamma}{c} \\ & \Rightarrow c\sin \alpha=a\sin \gamma \\ & \Rightarrow 18\cdot \frac{1}{2}=a\sin \gamma \\ & \Rightarrow a\sin \gamma=9 \\ & \Rightarrow \sin \gamma=\frac{9}{a} \\ & \Rightarrow \gamma =\arcsin \left (\frac{9}{a} \right ) \ \text{ with } \ a=\frac{293+ 9\sqrt{57}}{2} \ \text{ and } \ a=\frac{293- 9\sqrt{57}}{2}\end{align*} Is everything correct? Could I improve something? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

Since $\arcsin$ has range $[-\frac\pi 2,\frac\pi 2]$, doesn't that mean we're missing triangles with an oblique $\gamma$? (Wondering)

 

[mathmari]

Hey mathmari! (Smile)

Since $\arcsin$ has range $[-\frac\pi 2,\frac\pi 2]$, doesn't that mean we're missing triangles with an oblique $\gamma$? (Wondering)

Ahh (Thinking)

So do we have to do something else? (Wondering)

 

[I like Serena]

Ahh (Thinking)

So do we have to do something else? (Wondering)

It means that $\sin\gamma=\frac 9a \Rightarrow \gamma=\arcsin\frac 9a$ is incorrect.
The left hand side of the implication has more solutions than the right hand side.
So we have to extend the right hand side to include all solutions that the left hand side has.
We can use the restriction that $0<\gamma<\pi$ though, since it's given that $\gamma$ is an angle in a triangle. (Thinking)

 

[mathmari]

So we have to extend the right hand side to include all solutions that the left hand side has.
We can use the restriction that $0<\gamma<\pi$ though, since it's given that $\gamma$ is an angle in a triangle. (Thinking)

Do you mean that using that restriction we extend the right hand side? (Wondering)

 

[I like Serena]

Do you mean that using that restriction we extend the right hand side? (Wondering)

I mean that we should have:
$$\sin\gamma=\frac 9a\quad\Rightarrow\quad\gamma=\arcsin\frac 9a\lor\gamma=\pi-\arcsin\frac 9a$$
(Thinking)

 

[I like Serena]

$$\Rightarrow a^2=\frac{293\pm 9\sqrt{57}}{2} \\ \Rightarrow a=\frac{293\pm 9\sqrt{57}}{2}$$

Didn't we lose a $\sqrt{}$ here? (Wondering)

 

[I like Serena]

I think the solutions should look like:
\begin{tikzpicture}
\def\b{(9*sqrt(3)+sqrt(19))/2};
\def\ba{(9*sqrt(3)-sqrt(19))/2};
\coordinate (A) at (0,0);
\coordinate (B) at (18,0);
\coordinate (C) at ({\b*cos(30)},{\b*sin(30)});
\coordinate (C') at ({\ba*cos(30)},{\ba*sin(30)});
\coordinate (D) at (9,0);
\draw[orange, ultra thick] (A) -- (B) -- node[above] {a'} (C') node[above] {C'} -- node[below right] {b'} cycle;
\draw[orange, ultra thick] (C') -- node[below left] {5} (D);
\draw[blue, ultra thick] (A) node[below left] {A} -- node[below, xshift=30] {c} (B) node[below right] {B} --node[above right] {a} (C) node[above] {C} -- node[above left] {b} cycle;
\draw[blue, ultra thick] (C) -- node

{5} (D) node[below] {D};
\path[blue, ultra thick] (A) -- node[above] {9} (D) -- node[above] {9} (B);
\path[blue, ultra thick] (A) node[above right, xshift=18] {$\pi/6$};
\draw[blue, thick] (A) +(1.5,0) arc (0:30:1.5);
\end{tikzpicture}

Note that $\gamma > \frac\pi 2$ in both cases. (Thinking)​

 

[mathmari]

I mean that we should have:
$$\sin\gamma=\frac 9a\quad\Rightarrow\quad\gamma=\arcsin\frac 9a\lor\gamma=\pi-\arcsin\frac 9a$$
(Thinking)

So, in general it holds that $$\sin x=y\Rightarrow x=\arcsin y\lor x=\pi-\arcsin y$$ right? (Wondering)

Didn't we lose a $\sqrt{}$ here? (Wondering)

Oh yes (Tmi)

 

[I like Serena]

So, in general it holds that $$\sin x=y\Rightarrow x=\arcsin y\lor x=\pi-\arcsin y$$ right?

Generally, we have:
$$\sin x=y\quad\Rightarrow\quad x=\arcsin y + 2\pi k\lor x=\pi-\arcsin y + 2\pi k$$
If we know that $0<x<\pi$, we can conclude that $x=\arcsin y\lor x=\pi-\arcsin y$, and also that $y>0$. (Nerd)

 

[mathmari]

Generally, we have:
$$\sin x=y\quad\Rightarrow\quad x=\arcsin y + 2\pi k\lor x=\pi-\arcsin y + 2\pi k$$
If we know that $0<x<\pi$, we can conclude that $x=\arcsin y\lor x=\pi-\arcsin y$, and also that $y>0$. (Nerd)

Ahh ok! I see! Thank you so much! (Mmm)