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In a triangle $\Delta ABC$ let $D$ be the midpoint of $BC$. If angle $\angle ADB=45^{\circ}$ and angle $\angle ACD=30^{\circ}$ then find angle $\angle BAD$.

NOW this is a special case do we need a construction.

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Drop perpendicular from $A$ on extended $BC$ intersecting at $E$ to form right triangle $\triangle AEC$.

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Let $AC = b$

Then by basic trigonometry we get:

$EA=\frac{b}{2}$

$DC=EC-ED=\frac{\sqrt{3}-1}{2}b$

$EB=EC-BC=\frac{2-\sqrt{3}}{2}b$

$\angle EAB=\tan^{-1}\frac{EB}{EA}=\frac{\pi}{12}$

Therefore,

$$\angle BAD=\angle EAD-\angle EAB=\frac{\pi}{4}-\frac{\pi}{12}=\frac{\pi}{6}$$

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Let's denote $\angle BAD$ as $\alpha$. We have that $\angle ADB=180-(45+\alpha)$. If $\angle ADB=45$ and $\angle ACD=30$, then $\angle ADC=180-45$ and $\angle CDA=180-30-(180-45)=15$. Now, from the sine theorem:

$$ \frac{|CD|}{\sin(15)}=\frac{|AC|}{\sin(180-45)}$$ $$ \frac{|AC|}{\sin(180-(45+\alpha))}=\frac{|BD|}{\sin(\alpha)}=\frac{|CD|}{\sin(\alpha)}$$

From the first one, we have that:

$$ |AC|=\frac{\sin(180-45)}{\sin(15)}|CD|$$

Plugging this into the second one gives us:

$$ \frac{\sin(180-45)}{\sin(15)}|CD|\frac{1}{\sin(180-(45+\alpha))}=\frac{|CD|}{\sin(\alpha)} $$ $$ \frac{\sin(15)\sin(180-45)}{\sin(180-(45+\alpha))}=\frac{1}{\sin(\alpha)} $$ $$ \sin(\alpha)=\frac{\sin(180-(45+\alpha))}{\sin(15)\sin(180-45)}=\frac{\sin(45+\alpha)}{\sin(15)\sin(180-45)} $$

Can you take it from here?

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  • $\begingroup$ better do it fully $\endgroup$ – starunique2016 Jan 17 '17 at 11:23
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One can use a construction if one does not want to use trigonometry but only very simple but tricky methods.

Let point $B'$ be such that triangle $AB'C$ is equilateral and $B$ and $B'$ lie on the same side of line $AC.$ Let point $C'$ be such that line $AC'$ is orthogonal to $B'C$ and $AC' = AC = AB' = B'C$. By construction, $AC'$ is the orthgognal bisector of $B'C$ so $$B'C' = CC'\,\,\, \text{ and } \,\,\, \angle \, CAC' = B'AC' = 30^{\circ}$$ Therefore, triangles $CAC'$ and $B'AC'$ are congruent isosceles triangles with $30^{\circ}$ internal angle at vertex $A$ (because $AC'$ is the angle bisector of equlateral triangle $AB'C$).

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Now, by assumption, $\angle \, DAC = \angle \, ADB - \angle \, ACD = 45^{\circ} - 30^{\circ} = 15^{\circ}$. Hence, line $AD$ is the angle bisector in isosceles triagnle $CAC'$ through vertex $A$ and so $AD$ is also orthogonal bisectror of $CC'$ which means that $DC = DC'$. A direct angle chasing shows that $\angle \, CDC' = 90^{\circ}$. However, by assumption, $BD = DC = DC'$ so $$BC ' = CC' = B'C' \,\,\, \text{ and } \,\,\, \angle \, BC'C = 90^{\circ}$$ Hence triangle $BB'C'$ is isosceles and after a direct angle chasing, one sees that $\angle \, BC'B' = 60^{\circ}$. Thus $BB'C'$ is an equilateral triangle. This means that triangles $ABC'$ and $ABB'$ are congruent, because $BB' = BC'$, segment $AB$ is common and $AB'=AC'$ by construction. Therefore, $$\angle \, BAB' = \angle \, BAC' = \frac{1}{2} \, \angle \, B'AC' = \frac{1}{2} \, 30^{\circ} = 15^{\circ}$$ We can conclude that $$\angle \, BAD = \angle \, B'AC - \big(\angle \, BAB' + \angle DAC \big) = 60^{\circ} - (15^{\circ} + 15^{\circ}) = 30^{\circ}$$

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  • $\begingroup$ good one brother $\endgroup$ – starunique2016 Jan 17 '17 at 11:23

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