# Find an angle of a triangle on a larger triangle which cut through its midpoint

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 $\angle BAC$.

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?

• I think that this can be solved suing just F and Z angles. Draw a line parallel to $AB$ that goes through $C$. – stuart stevenson Dec 20 '17 at 18:57
• @stuartstevenson Ok....but....can you please go through my method?....i want to know why it is not working. – ami_ba Dec 20 '17 at 19:01
• i think your second equation is wrong! – Dr. Sonnhard Graubner Dec 20 '17 at 19:08
• I don't think there's a problem with it. – stuart stevenson Dec 20 '17 at 19:10
• but i think so, you can not use $$45^{\circ}$$ AND $$\theta$$ in the triangle $$\Delta ADC$$ – Dr. Sonnhard Graubner Dec 20 '17 at 19:14

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 $\cot\theta=\sqrt{3}$, $\theta=30°$

• Perfect....thank you☺️☺️ – ami_ba Dec 21 '17 at 3:35

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 $$\alpha$$

Never Underestimate the Power of Euclidean Geometry

• 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 $$\angle BAC = 45°$$.

$$\blacksquare$$