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In triangle $\triangle BAC$ with $\angle ABC = 30\deg$. $D$ is the midpoint of $BC$. We join $A$ and $D$ and $\angle CDA = 45 \deg$. Find $\angle BAC$.

the image representation of the problem

On applying Sine rule, $$\frac{2x}{\sin {(15+\theta)}}=\frac{AC}{\sin 30}$$ and also $$\frac{x}{\sin \theta}=\frac{AC}{\sin 45}$$
Where $x$ is $CD$ or $DB$ and $\theta$ is $\angle CAD$.

But solving this gives $$\frac{\sin {(15+\theta)}}{\sin \theta}=\sqrt 2$$

Is this correct?

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  • $\begingroup$ I think that this can be solved suing just F and Z angles. Draw a line parallel to $AB$ that goes through $C$. $\endgroup$ – stuart stevenson Dec 20 '17 at 18:57
  • $\begingroup$ @stuartstevenson Ok....but....can you please go through my method?....i want to know why it is not working. $\endgroup$ – ami_ba Dec 20 '17 at 19:01
  • $\begingroup$ i think your second equation is wrong! $\endgroup$ – Dr. Sonnhard Graubner Dec 20 '17 at 19:08
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    $\begingroup$ I don't think there's a problem with it. $\endgroup$ – stuart stevenson Dec 20 '17 at 19:10
  • $\begingroup$ but i think so, you can not use $$45^{\circ}$$ AND $$\theta$$ in the triangle $$\Delta ADC$$ $\endgroup$ – Dr. Sonnhard Graubner Dec 20 '17 at 19:14
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Your reasoning looks good to me. Using your second equation, $$AC=\frac{x}{\sqrt{2}\sin{\theta}}$$ Now substituting $AC$ in the first equation, $$\frac{2x}{\sin{(15+\theta)}}=\frac{2x}{\sqrt{2}\sin{\theta}}$$ or $$\sin{(15+\theta)}=\sqrt{2}\sin{\theta}$$ Using trig identity, $$\cos15\sin\theta+\sin15\cos\theta=\sqrt{2}\sin{\theta}$$ Dividing by $\sin\theta$ we get $$\cot\theta=\frac{\sqrt{2}-\cos15}{\sin15}$$ Knowing that $\sin15=\frac{\sqrt{3}-1}{2\sqrt{2}}$, $\cos15=\frac{\sqrt{3}+1}{2\sqrt{2}}$ we get $\cot\theta=\sqrt{3}$, $\theta=30°$

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  • $\begingroup$ Perfect....thank you☺️☺️ $\endgroup$ – ami_ba Dec 21 '17 at 3:35
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you will Need three equations $$a^2=b^2+c^2-2bc\cos(\alpha)$$ $$a=\frac{b\sin(\alpha)}{\sin(30^{\circ})}$$ $$c=\frac{b\sin(150^{\circ})}{\sin(30^{\circ})}$$ then you Can divide by $b^2$ and you will get only an equation for $$\alpha$$

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Never Underestimate the Power of Euclidean Geometry

enter image description here

  • By Exterior Angle Theorem $\angle DAB = 15°$.
  • Draw from $C$ the line perpendicular to $AB$, intersecting $AB$ in $E$. We then have $\angle ECB = 60°$ and thus $\triangle EBC$ is half of an equilater triangle and $CE \cong \frac{BC}{2} \cong CD \cong BD$.
  • Now connect $E$ with $D$. $\triangle CED$ is isosceles, but having $\angle ECB = 60°$, it is also equilateral and thus $ED \cong CE$, and $\angle EDA = \angle DAB = 15°$.
  • So $\triangle AED$ is isosceles and we get also $AE \cong CE$.
  • We conclude that $\triangle ACE$ is isosceles and right-angled. Therefore $\angle BAC = 45°$.

$\blacksquare$

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