# Solve a triangle given one angle of the median, and a side and an angle of its external triangle

This is a modified version of this question, and as such I'm using similar wording and visuals. Given:

• $$α_2$$, one of the two angles which the vertex $$A$$ is split by the median $$m$$;
• $$\overline{AD}$$, the length of the segment of a triangle external to $$ABC$$;
• $$\overline{BM}$$, the length of half of the side split by the median;
• $$θ$$, the angle opposite to the shared side $$\overline{AB}$$;

Find the value of the angles $$γ$$ and $$b_2$$.

Similarly to the original question, the proportion $$\frac{\sin{α_1}}{\sin{α_2}}=\frac{\sin{β_1}}{\sin{γ}}$$ holds true. And it's still true that $$β_1 = 180 - α_1 - α_2 - γ$$. However, unlike the original scenario, we don't know the value of $$α_1$$.

With the external angle theorem, we know that $$α_1 = θ + β_2 - α_2$$, but now we have the $$β_2$$ variable to resolve.

I figured we could use the law of sines to establish $$\frac{\sin{β_2}}{\overline{AD}}=\frac{\sin{θ}}{\overline{AB}}$$, but after a few variable swaps the problem ultimately seems to loop back to $$α_1$$.

I feel that this problem is solvable, as the given parameters uniquely identify the two triangles. But I can't find out how to take it further than I have.

Let $$x:=\gamma, y:=\theta+\beta_2, p:=\overline{AD}$$ and $$q:=\overline{BM}=\overline{CM}$$.

Then, one can write $$\angle{ABC}=180-x-y\qquad\text{and}\qquad \angle{BAM}=y-\alpha_2$$

Applying the law of sines to $$\triangle{ABC}$$, one has $$\overline{AC}=\frac{2q\sin(180-x-y)}{\sin(y-\alpha_2+\alpha_2)}\tag1$$

Applying the law of sines to $$\triangle{AMC}$$, one has $$\overline{AM}=\frac{q\sin x}{\sin\alpha_2}\tag2$$

Applying the law of sines to $$\triangle{AMB}$$, one has $$\overline{AM}=\frac{q\sin(180-x-y)}{\sin(y-\alpha_2)}\tag3$$ Applying the law of sines to $$\triangle{DBC}$$, one has $$p+\overline{AC}=\frac{2q\sin(\beta_2+180-x-y)}{\sin\theta}\tag4$$

It follows from $$(1)(4)$$ that $$p+\frac{2q\sin(180-x-y)}{\sin(y-\alpha_2+\alpha_2)}=\frac{2q\sin(\beta_2+180-x-y)}{\sin\theta}$$ i.e. $$p+\frac{2q\sin(x+y)}{\sin y}=\frac{2q\sin(x+\theta)}{\sin\theta}$$ Multiplying the both sides by $$\sin y\sin\theta$$ gives $$p\sin y\sin\theta+2q\sin(x+y)\sin\theta=2q\sin(x+\theta)\sin y$$ i.e. $$p\sin y\sin\theta+2q\sin\theta\sin x\cos y+2q\sin\theta\cos x\sin y=2q\sin y\sin x\cos\theta+2q\sin y\cos x\sin\theta$$ i.e. $$(p\sin\theta-2q\sin x\cos\theta)\sin y+2q\sin\theta\sin x\cos y=0\tag5$$

It follows from $$(2)(3)$$ that $$\frac{q\sin x}{\sin\alpha_2}=\frac{q\sin(180-x-y)}{\sin(y-\alpha_2)}$$ i.e. $$\frac{\sin x}{\sin\alpha_2}=\frac{\sin(x+y)}{\sin(y-\alpha_2)}$$ Multiplying the both sides by $$\sin\alpha_2\sin(y-\alpha_2)$$ gives $$\sin x\sin(y-\alpha_2)=\sin\alpha_2\sin(x+y)$$ i.e. $$\sin x\sin y\cos\alpha_2-\sin x\cos y\sin\alpha_2=\sin\alpha_2\sin x\cos y+\sin\alpha_2\cos x\sin y$$ i.e. $$(\sin\alpha_2\cos x-\sin x\cos\alpha_2)\sin y+2\sin\alpha_2\sin x\cos y=0\tag6$$

Now, $$\sin\alpha_2\times (5)-q\sin\theta\times (6)$$ gives $$\sin\alpha_2(p\sin\theta-2q\sin x\cos\theta)\sin y-q\sin\theta(\sin\alpha_2\cos x-\sin x\cos\alpha_2)\sin y=0$$ Dividing the both sides by $$\sin y$$ gives $$S\cos x+T\sin x=U$$ where $$\begin{cases}S=q\sin\theta\sin\alpha_2\gt 0 \\T=2q\sin\alpha_2\cos\theta-q\sin\theta\cos\alpha_2 \\ U=p\sin\alpha_2\sin\theta\gt 0\end{cases}$$ So, we can write $$\sqrt{S^2+T^2}\cos(x-V)=U$$ where $$V=\arcsin\bigg(\frac{T}{\sqrt{S^2+T^2}}\bigg)$$ which satisfies $$-90\lt V\lt 90$$.

So, one gets $$x=V\pm\arccos\bigg(\frac{U}{\sqrt{S^2+T^2}}\bigg)$$ where one has to choose $$x$$ satisfying $$0\lt x\lt 180$$.

It follows from $$(6)$$ that $$y=\begin{cases}90&\text{if x=\alpha_2} \\\\\arctan W_x&\text{if x\not=\alpha_2 and W_x\gt 0} \\\\180+\arctan W_x&\text{if x\not=\alpha_2 and W_x\lt 0}\end{cases}$$ where $$W_x=\frac{-2\sin\alpha_2\sin x}{\sin(\alpha_2-x)}$$

In conclusion, one gets \color{red}{\begin{align}\gamma&=V\pm\arccos\bigg(\frac{U}{\sqrt{S^2+T^2}}\bigg) \\\\\beta_2&=\begin{cases}90-\theta&\text{if \gamma=\alpha_2} \\\\-\theta+\arctan W_{\gamma}&\text{if \gamma\not=\alpha_2 and W_{\gamma}\gt 0}\\\\-\theta+180+\arctan W_{\gamma}&\text{if \gamma\not=\alpha_2 and W_{\gamma}\lt 0}\end{cases}\end{align}} where $$\begin{cases}0\lt\gamma\lt 180 \\\max(0,\alpha_2-\theta)\lt \beta_2\lt 180-\theta-\gamma \\S=\overline{BM}\sin\theta\sin\alpha_2 \\T=2\overline{BM}\sin\alpha_2\cos\theta-\overline{BM}\sin\theta\cos\alpha_2 \\ U=\overline{AD}\sin\alpha_2\sin\theta \\\displaystyle V=\arcsin\bigg(\frac{T}{\sqrt{S^2+T^2}}\bigg) \\\displaystyle W_{\gamma}=\frac{-2\sin\alpha_2\sin\gamma}{\sin(\alpha_2-\gamma)}\end{cases}$$

By the law of sines in $$\triangle ABM$$ and $$\triangle ABD$$, we have

$$\begin{multline*} BM \sin (α_2 + γ) = AB \sin α_1 = \frac{AD\sin θ \sin α_1}{\sin β_2} = \frac{AD \sin θ \sin (β_2 + θ - α_2)}{\sin β_2} \\ = AD \sin θ \cos (θ - α₂) + AD \sin θ \sin (θ - α₂) \cot β_2. \end{multline*}$$

Meanwhile, by the law of sines in $$\triangle ABM$$ and $$\triangle ACM$$, we have

$$\begin{multline*} 0 = \frac{\sin α_1 \sin γ - \sin α_2 \sin β_1}{\sin β_2} = \frac{\sin (β_2 + θ - α_2) \sin γ - \sin α_2 \sin (β_2 + θ + γ)}{\sin β_2} \\ = \cos (θ - α_2) \sin γ - \sin α_2 \cos (θ + γ) + [\sin (θ - α_2) \sin γ - \sin α_2 \sin (θ + γ)] \cot β_2. \end{multline*}$$

Subtracting $$AD \sin θ \sin (θ - α_2)$$ times the second equation from $$[\sin (θ - α_2) \sin γ - \sin α_2 \sin (θ + γ)]$$ times the first equation eliminates $$β_2$$:

$$\begin{multline*} [\sin (θ - α_2) \sin γ - \sin α_2 \sin (θ + γ)] BM \sin (α_2 + γ) \\ = [\sin (θ - α_2) \sin γ - \sin α_2 \sin (θ + γ)] AD \sin θ \cos (θ - α_2) \\ - [\cos (θ - α_2) \sin γ - \sin α_2 \cos (θ + γ)] AD \sin θ \sin (θ - α_2). \end{multline*}$$

If we let $$t = \tan \frac γ2$$, so that $$\cos γ = \frac{1 - t^2}{1 + t^2}$$, $$\sin γ = \frac{2t}{1 + t^2}$$, we can expand the above into a quartic equation in $$t$$, and solve it using the quartic formula. We can then solve for $$γ$$ and use one of the first two equations to solve for $$β_2$$.

To better highlight the unknown terms, let us set $$\alpha_1=x$$, $$\beta_1=y$$, $$\gamma=z$$. The known quantities are $$\alpha_2$$, $$AD$$, $$BM$$, $$\theta$$.

The first two equations, already reported in the OP, are:

$$\frac{\sin{x}}{\sin{α_2}}=\frac{\sin{y}}{\sin{z}} \tag{1}$$ $$y = 180 - x - α_2 - z \tag{2}$$

As we have $$3$$ unknowns, we need a third equation.

From the triangle $$\triangle{ABD}$$ we have:

$$\frac{AB}{\sin \theta}=\frac{AD}{\sin \beta_2}$$

Here we can try to express $$AB$$ and $$\beta_2$$ as functions of $$x,y,z$$ and the other known terms. Considering the triangle $$\triangle{ABC}$$, we have

$$AB=\frac{2BM}{\sin(x+\alpha_2)} \,\sin z$$

Also, by the exterior angle theorem, $$\beta_2=x+\alpha_2-\theta$$

Substituting, we obtain our third equation, which complete the system:

$$\frac{2\,BM \, \sin z }{\sin(x+\alpha_2)} \,= \frac{AD \, \sin \theta }{\sin (x+\alpha_2-\theta)} \tag{3}$$

The system can then be solved by the usual methods. Once we obtained $$x,y,z$$, we can also calculate $$\beta_2$$ using the formula above.

To provide an example of how the system works: let us consider the very simple case where the triangle $$\triangle{ABC}$$ is right and isosceles with hypotenuse equal to $$2$$, and the triangle $$\triangle{ABD}$$ is right with $$\theta=60°$$. In this case, the known terms are $$\alpha _2=\pi/4$$, $$\gamma=\pi/3$$, $$BM=1$$, and $$AD=1/\sqrt{6}$$. By construction, this case corresponds to the trivial case where $$x=y=z=\pi/4$$ and $$\beta_2=\pi/6$$.

As expected, the system provides the correct $$x,y,z$$ solutions, as shown here. From this, we also easily get $$\beta_2=\pi/4+\pi/4-\pi/3$$ $$=\pi/6$$.

Changing-up notation slightly, I'll use $$\alpha$$, $$\beta$$, $$\delta$$ for OP's $$\alpha_2$$, $$\beta_2$$, $$\theta$$. Also, let $$N$$ and $$P$$ be the feet of the perpendiculars from $$A$$ to $$\overline{BD}$$ and $$M$$ to $$\overline{AC}$$, and let $$\square DPQR$$ be a rectangle, with $$B$$ on $$\overline{QR}$$. Finally, define $$a := |BM|=|MC| \qquad d := |AD| \qquad p := |MP|=|QM|$$ I'll also rotate the figure to make $$\overline{DC}$$ horizontal:

For the configuration shown, straightforward right-triangle trigonometry gives the (signed) lengths of various segments. Equating opposite sides of $$\square DPQR$$, and invoking the Pythagorean Theorem in $$\triangle CMP$$, gives these three equations: \begin{align} 2p &= d(\cot\beta+\cot\delta) \sin^2\delta \quad\left(= d\frac{\cot\beta+\cot\delta}{1+\cot^2\delta}\right) \tag1 \\ d+p \cot\alpha &= 2p\cot\delta+p\cot\gamma \tag2 \\[0.5em] a^2 &= p^2 + p^2 \cot^2\gamma \tag3 \end{align} Equations $$(1)$$ and $$(2)$$ give expressions involving our target angles. Defining $$u := \cot\alpha \qquad v := \cot\delta \qquad \lambda := u-2v$$ we have $$\cot\beta = \frac{2p(1+v^2)-dv}{d} \qquad\qquad \cot\gamma = \frac{d+p\lambda}{p}\tag4$$ Substituting $$\cot\gamma$$ into $$(3)$$ gives a quadratic in $$p$$ that we solve as $$p = \frac{-d\lambda\pm\sqrt{a^2(1+\lambda^2)-d^2}}{1+\lambda^2} \tag5$$ discarding non-positive root(s). Substituting from $$(5)$$ back into $$(4)$$ determines $$\beta$$ and $$\gamma$$. $$\square$$

(Considering whether there can be two positive roots $$p$$ —hence, two overall solutions— is left as an exercise to the reader.)

As something of a sanity check ... Consider the isosceles right triangle example in @Anatoly's answer.

We have $$\alpha=45^\circ \qquad \delta=60^\circ \qquad a = 1 \qquad |AB|=\sqrt{2} \qquad d = \frac{\sqrt{6}}{3}$$ so that $$u = 1 \qquad v = \frac{\sqrt{3}}3 \qquad \lambda = \frac13(3-2\sqrt{3}) \qquad\to\qquad p =\frac{\sqrt{2}}{2}$$ (Note: Reducing $$(5)$$ to $$\sqrt{2}/2$$ is a non-trivial exercise!) Then, $$\cot\beta = \sqrt{3} \qquad \cot\gamma = 1 \qquad\to\qquad \beta=30^\circ \qquad \gamma=45^\circ$$ as expected. $$\square$$