Could you help me approach this problem using sine law?

Question

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

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

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

ABC is isosceles, but you do not tell us which of the two sides are congruent. Can we assume $AB$ and $AC$ are congruent? — Noble Mushtak, Nov 30 '18 at 23:00

score 1 · Answer 1 · answered Nov 30 '18 at 23:25

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.

score 1 · Answer 2 · answered Nov 30 '18 at 23:39

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.

user · Answer 3 · 2018-12-01T08:23:59.020

0

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

$\frac{a}{\sin \alpha}=\frac{b}{\sin 80}$
$\frac{a}{\sin 60}=\frac{x}{\sin 80}=\frac{y}{\sin 40}$
$\frac{x}{\sin \alpha}=\frac{b}{\sin 120}=\frac{b-y}{\sin 40}$

edited Dec 01 '18 at 08:23

answered Nov 30 '18 at 23:05

user

154,566

Where does the 40 come from? – Noble Mushtak Nov 30 '18 at 23:10
@NobleMushtak $\angle BDC=60°$ – user Nov 30 '18 at 23:11
That does not really make it obvious how you got the $40^\circ$ using Law of Sines. – Noble Mushtak Nov 30 '18 at 23:26
Also, the second equation should have $\sin \alpha$, not $\sin A$. $\alpha$ is the measure of $\angle ABD$ and $\angle BDC$. – Noble Mushtak Nov 30 '18 at 23:27
@NobleMushtak The triangle is isosceles, therefore $\angle B=\angle C$ and since $B/2+C+60=180 \implies B=C=80$. – user Nov 30 '18 at 23:28
@NobleMushtak For the notation it is really a minor issue, I've just used $A$ to indicate the angle, it seems totally clear from the context. – user Nov 30 '18 at 23:29
No, it is not a minor issue, because the equations in your answer imply $\angle A=40^\circ$, when it is actually not. Instead, $\alpha=40^\circ$ and $A=60^\circ-\alpha=20^\circ$. – Noble Mushtak Nov 30 '18 at 23:30
@NobleMushtak Ah ok thanks! I fix that. – user Nov 30 '18 at 23:32

Could you help me approach this problem using sine law?

3 Answers3