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Given that ABC is an isosceles triangle, $[BD]$ is angle bisector, $\angle BDA = 120^\circ$. Evaluate the degree of $\angle A $

Could you help me approach this problem using sine law?

Here's my attempt:

From the angle bisector theorem, we know that

$$\dfrac{|AB|}{|BC|} = \dfrac{|AD|}{|DC|} $$

In $\triangle ADB$, let's call same angles $\alpha$ and we have that

$$\dfrac{|AB|}{\sin (120)} = \dfrac{|AD|}{\sin(\alpha )} \implies \dfrac{|AB|}{|AD|} = \dfrac{\sin (120)}{\sin (\alpha )} $$

This also equals

$$\dfrac{|AB|}{|AD|} = \dfrac{\sin (120)}{\sin (\alpha )} = \dfrac{|BC|}{|DC|}$$

Now $\angle A 180-120-\alpha = 60-\alpha $, then

$$\dfrac{|DB|}{\sin (x)} = \dfrac{|AB|}{\sin (120)} \implies \dfrac{|DB|}{|AB|} = \dfrac{\sin(60-\alpha )}{\sin (120)} $$

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    $\begingroup$ Evauate the degree of? angle in A I suppose $\endgroup$ – user Nov 30 '18 at 22:48
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    $\begingroup$ @gimusi Yes, that's right. However, I want to use sine law. $\endgroup$ – Hamilton Nov 30 '18 at 22:53
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    $\begingroup$ ABC is isosceles, but you do not tell us which of the two sides are congruent. Can we assume $AB$ and $AC$ are congruent? $\endgroup$ – Noble Mushtak Nov 30 '18 at 23:00

I think Law of Sines is ill-suited for this problem. However, in order to understand this, you first need to see the answer. First, $AB=AC$ since triangle $ABC$ is isosceles. Therefore:

$$\angle ABC=\angle ACB$$

Also, $BC$ is an angle bisector, so:

$$\angle ABD=\frac 1 2\angle ABC$$

From triangle $ABD$, we have:

$$\angle A+\frac 1 2\angle ABC+120^\circ=180^\circ$$

From triangle $ABC$, we have:

$$\angle A+2\angle ABC=180^\circ$$

Subtract the second equation by the first:

$$\frac 3 2\angle ABC-120^\circ=0\rightarrow \angle ABC=80^\circ$$

Substitute back into the second equation:

$$\angle A+160^\circ=180^\circ\rightarrow\angle A=20^\circ$$

Thus, our final answer is $20^\circ$. This means:

$$\sin A=\frac i 2\left(\sqrt[3]{\frac{-1-\sqrt{-3}}{2}}-\sqrt[3]{\frac{-1+\sqrt{-3}}{2}}\right)$$

I think it would be very hard to derive this complex expression from using Law of Sines in order to solve for A, which is why Law of Sines is not a good way to solve this rather simple angle problem.

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Let $x$ be the measure of $\angle ABD$ and $y$ be the measure of $\angle BAC$

$x+y+ 120 = 180\\ 4x + y = 180$

And solve the system of equations.

Regrading law of sines.. that might be useful you knew more side lengths. Right now all you know is that the triangle is isosceles.

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We have that since $\angle BDC=60° \implies \angle ACB=\angle ABC=80*$ and $\angle BAC=20°$.

To find the result by law of sine let

  • $BC=a $
  • $AB=AC=b$
  • $BD=x$
  • $DC=y$
  • $\angle BAC=\alpha$

then we have to solve the following systems of $5$ equations in $5$ unknowns

  • $\frac{a}{\sin \alpha}=\frac{b}{\sin 80}$
  • $\frac{a}{\sin 60}=\frac{x}{\sin 80}=\frac{y}{\sin 40}$
  • $\frac{x}{\sin \alpha}=\frac{b}{\sin 120}=\frac{b-y}{\sin 40}$
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  • $\begingroup$ Where does the 40 come from? $\endgroup$ – Noble Mushtak Nov 30 '18 at 23:10
  • $\begingroup$ @NobleMushtak $\angle BDC=60°$ $\endgroup$ – user Nov 30 '18 at 23:11
  • $\begingroup$ That does not really make it obvious how you got the $40^\circ$ using Law of Sines. $\endgroup$ – Noble Mushtak Nov 30 '18 at 23:26
  • $\begingroup$ Also, the second equation should have $\sin \alpha$, not $\sin A$. $\alpha$ is the measure of $\angle ABD$ and $\angle BDC$. $\endgroup$ – Noble Mushtak Nov 30 '18 at 23:27
  • $\begingroup$ @NobleMushtak The triangle is isosceles, therefore $\angle B=\angle C$ and since $B/2+C+60=180 \implies B=C=80$. $\endgroup$ – user Nov 30 '18 at 23:28

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