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.
[Math] In a triangle $ABC$ let $D$ be the midpoint of $BC$ . If $\angle ADB=45^\circ$ and $\angle ACD=30^\circ$ then find $\angle BAD$
geometry
Related Solutions
When I see that $\angle CAB$ is partitioned as $24^\circ$ and $18^\circ$, what angle is left to make an $60^\circ$ angle (which is for making an equilateral triangle, which gives us more equal sides)? After doing that, we see that $\angle ECB = 36^\circ$ and this makes $\angle ABE = 30^\circ$. Now, notice that $\Delta ADB$ is congruent to $\Delta AEB$. So we have $|AD| = |AE| = |AC|$. So the answer is $78^\circ$.
reflex $\angle ADC=360^{\circ}-146^{\circ}=214^{\circ}=2\times 107^{\circ}=2\times \angle ABC$
Since the angle subtended at the center of a circle is double the angle subtended at the circumference, above proves that our constructed circle also passes through the point $B$. So, $|DA|=|DB|=|DC|$ which in turn means that $\angle ABD=\angle BAD=\angle DAC+\angle BAC=47^{\circ}$.
Related Question
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Best Answer
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?