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$\triangle ABC$ is an isosceles triangle with $AB=BC$ and $\angle ABD=60^{\circ}$, $\angle DBC=20^{\circ}$ and $\angle DCB=10^{\circ}$. Find $\angle BDA$.

My approach: Let $\angle BDA=x$. Let $AB=BC=p$. Applying sine law in $\triangle ADB$, $\dfrac{p}{\sin x}=\dfrac{BD}{\sin (60+x)}$. Applying sine law in $\triangle BDC$, $\dfrac{p}{\sin150^{\circ}}=\dfrac{BD}{\sin 10^{\circ}}$. Using the two equations, we get $\dfrac{1}{2\sin 10^\circ}=\dfrac{\sin x}{\sin (60^\circ +x)} \implies 2\sin 10^\circ=\dfrac{\sqrt{3}}{2}\cot x + \dfrac{1}{2} \\ \implies x = \text{arccot} \left(\dfrac{4\sin 10^\circ-1}{\sqrt{3}}\right)$.

Now I am stuck. I know that the answer is $100^\circ$ but no matter how hard I try I cannot seem to simplify it any further. Please help. If anybody has a better solution (involving simple Euclidean Geometry), I would be grateful if you provide it too.

Edit: I am extremely sorry. The original problem was when $AB=BC$. Sorry for the inconvenience caused. I have rectified my mistake. Also, I have changed the answer to $100 ^\circ$.

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  • $\begingroup$ Care to explain? I don't see how. $\angle BAC=50^\circ$ and $\angle BAD=40^\circ$ if $\angle BDA=80^\circ$. $\endgroup$ – Popular Power Apr 25 '20 at 7:41
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After erecting an equilateral triangle ACE

$\angle ABC=\angle ABD+\angle DBC=80^\circ$.

\begin{align*} AB&=BC\\ \implies \angle CAB&=\angle BCA=(180^\circ-\angle ABC)/2=50^\circ. \end{align*}

Erect an equilateral triangle $ACE$ on base $AC$. Then $\triangle$s $ABE, CBE$ are congruent in opposite sense because $AB=CB$, $AE=CE$ and $BE$ is common. Thus $$\angle AEB=\angle BEC=30^\circ.$$

$$\angle CDB=180^\circ-\angle DBC-\angle BCD=150^\circ.$$ Thus quadrilateral $BDCE$ is cyclic because its angles $D$ and $E$ are supplementary. Thus $$\angle DEC=\angle DBC=20^\circ.$$

\begin{align*} \angle ECB&=\angle ECA-\angle BCA=10^\circ\\ \implies \angle ECD&=\angle ECB+\angle BCD=20^\circ=\angle DEC. \end{align*}

Thus triangle $CED$ is isosceles on base $CE$, so $CD=DE$. Thus $\triangle$s $ACD, AED$ are congruent in opposite sense because $AC=AE$, $CD=ED$ and $AD$ is common. Thus

\begin{align*} \angle CAD&=\angle DAE=30^\circ\\ \angle BAE&=\angle CAE-\angle CAB=10^\circ\\ \implies \angle DAB&=\angle DAE-\angle BAE=20^\circ\\ \implies \angle BDA&=180^\circ-\angle DAB-\angle ABD=100^\circ. \end{align*}

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Let $E$ be the circumcenter of $BCD$. Then $\angle BED=2\angle BCD=20^\circ$ and $\angle DEC =2\angle DBC =40^\circ$. Hence $\angle BEC=60^\circ$. This and $BE=EC$ shows that $BEC$ is equilateral. So $BC=BE$ and $\angle CBE=60^\circ$. By assumption $AB=BC$, so $AB=BE$ and $$\angle BEA = 90^\circ -\frac 12 \angle ABE =90^\circ -\frac 12 \cdot 140^\circ =20^\circ =\angle BED.$$ Therefore $A,D,E$ are collinear and we find $$\angle BDA =180^\circ -\angle EDB = \angle BED+\angle DBE= 20^\circ+80^\circ =100^\circ.$$

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Continue to simplify

$$\begin{align} \cot x & =\frac{4\sin 10-1}{\sqrt{3}} =\frac{(2\sin 10-\frac12)\cos10}{\frac{\sqrt{3}}2\cos10} \\ & =\frac{\sin 20-\cos60\cos10}{\cos10\sin60} =\frac{2\cos 70-2\cos60\cos10}{\cot10\cdot2\sin10\sin60} \\ & =\frac{\cos70-\cos50}{\cot10\cdot(\cos50-\cos70)} =-\cot80=\cot100 \end{align}$$

Thus, $x=100^\circ$.

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Assuming $AB=BC$ is what you intended, your calculation is correct. Notice that $\frac{4 \sin 10^\circ - 1}{\sqrt 3}$ is negative, and in fact the arccot of this value is $-80^\circ$. How can the angle be negative?! Recall that $x$ has to be an obtuse angle, so you should add $180^\circ$ to $-80^\circ$, obtaining $100^\circ$. You can confirm that $x=100^\circ$ also satisfies the equation you obtained.

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  • $\begingroup$ Can you please explain why you added $180^\circ$ instead of $360^\circ$? $\endgroup$ – Popular Power Apr 25 '20 at 17:22
  • $\begingroup$ @PopularPower The cotangent and tangent functions have period $180^\circ$. In other words, $\cot(x) = \cot(x+180)$. It is true that $360^\circ + x$ also satisfies your equation, but that angle is impossible in a triangle. $\endgroup$ – grand_chat Apr 25 '20 at 19:05
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It you are looking for a "clever" way to solve the obtained trigonometric equation, the following trick is often useful in similar problems:

Let $x $ satisfy the equation: $$ \frac {\sin (x)}{\sin (C-x)}=\frac {\sin (A)}{\sin (C-A)},\quad 0<x,A <C <\pi.\tag1 $$ Then $$ x=A.\tag2$$

Applying this to your problem one obtains:

$$\frac {\sin (x)}{\sin (120^\circ-x)}=\frac1{2\sin 10^\circ} =\frac{\cos 10^\circ}{\sin 20^\circ}=\frac{\sin 100^\circ}{\sin 20^\circ}\implies x=100^\circ. $$


Proof of $(1)\implies (2) $: $$\begin{align} &\frac {\sin x}{\sin (C-x)}=\frac {\sin A}{\sin (C-A)}\\ &\iff \sin x\,(\sin C \cos A-\cos C\sin A)=\sin A\,(\sin C \cos x-\cos C\sin x)\\ &\iff \sin C\,(\sin x\cos A-\cos x \sin A)=0\\ &\iff\sin C\sin(x-A)=0\stackrel{0<x,A <C <\pi}\implies x=A. \end{align} $$

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  • $\begingroup$ Could you provide a proof of the same? $\endgroup$ – Popular Power Apr 25 '20 at 16:39
  • $\begingroup$ @PopularPower I have added a proof. $\endgroup$ – user Apr 25 '20 at 18:37
  • $\begingroup$ Thank you so much! $\endgroup$ – Popular Power Apr 26 '20 at 3:09
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Although not as satisfying as a purely geometrical solution, the most direct method is to apply the Trigonometric Form of Ceva's Theorem : $$\frac{\sin\alpha}{\sin(A-\alpha)}.\frac{\sin\beta}{\sin(B-\beta)}.\frac{\sin\gamma}{\sin(C-\gamma)}=1$$ where $A, B, C$ are the angles of the triangle which are split by the concurrent cevians into angles $\alpha, A-\alpha, \beta, B-\beta, \gamma, C-\gamma$ in order round the triangle.

The resulting equation of the form $$R\sin\alpha=\sin(A-\alpha)$$ has the solution $$\tan\alpha=\frac{\sin A}{R+\cos A}$$ In your problem $$R=\frac{\sin40^{\circ}}{\sin10^{\circ}}.\frac{\sin20^{\circ}}{\sin60^{\circ}}=1.4619022$$ $$\tan\alpha=\frac{\sin50^{\circ}}{1.4619022+\cos50^{\circ}}=0.36397$$ $$\alpha=20^{\circ}$$ $$\angle BDA = 180^{\circ}-60^{\circ}-\alpha=100^{\circ}$$

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  • $\begingroup$ Thanks! Did not think of that one $\endgroup$ – Popular Power May 10 '20 at 15:56

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