# Geometry Problem on $\triangle ABC$ and Angle Chasing $$\triangle ABC$$ 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 $$100 ^\circ$$.

• Care to explain? I don't see how. $\angle BAC=50^\circ$ and $\angle BAD=40^\circ$ if $\angle BDA=80^\circ$. – Popular Power Apr 25 '20 at 7:41 $$\angle ABC=\angle ABD+\angle DBC=80^\circ$$.

\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*} 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 $$\angle BDA =180^\circ -\angle EDB = \angle BED+\angle DBE= 20^\circ+80^\circ =100^\circ.$$

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, $$x=100^\circ$$.

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.

• Can you please explain why you added $180^\circ$ instead of $360^\circ$? – Popular Power Apr 25 '20 at 17:22
• @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. – grand_chat Apr 25 '20 at 19:05

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: $$\frac {\sin (x)}{\sin (C-x)}=\frac {\sin (A)}{\sin (C-A)},\quad 0 Then $$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

• Could you provide a proof of the same? – Popular Power Apr 25 '20 at 16:39
• @PopularPower I have added a proof. – user Apr 25 '20 at 18:37
• Thank you so much! – Popular Power Apr 26 '20 at 3:09

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 $$R=\frac{\sin40^{\circ}}{\sin10^{\circ}}.\frac{\sin20^{\circ}}{\sin60^{\circ}}=1.4619022$$ $$\tan\alpha=\frac{\sin50^{\circ}}{1.4619022+\cos50^{\circ}}=0.36397$$ $$\alpha=20^{\circ}$$ $$\angle BDA = 180^{\circ}-60^{\circ}-\alpha=100^{\circ}$$

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