Finding angle in a given triangle.

Question

In the picture above:

• $\overset{\Delta}{ACD}$ is a triangle.
• $B$ is a point on $[CD]$.
• $m(\widehat{ABC})=140^\circ$
• $|AB|=|BC|$.
• $|AC|=|BD|$.
• What is $\color{red}{m(\widehat{ADB})}$?

There is probably a short answer, but i can't find it.

[Answer is $\color{red}{30^\circ}$.]

Answer 1

Obviously, $\angle BAC = \angle BCA = 20^{\circ}$, and $\angle DBA = 40^{\circ}$.

Choose point $E$ such that $\triangle EAB$ forms an equilateral triangle.

Since $\angle DBA = 40^{\circ}$, and $\angle ABE = 60^{\circ}$, then $\angle DBE = 20^{\circ}$.

Now notice: $\overline {AC} = \overline {DB}$; $\angle ACB = \angle DBE = 20^{\circ}$; and $\overline {CB} = \overline {BE}$. We have side-angle-side equivalence, so, if we draw in segment $DE$, we can say that $\triangle ACB \cong \triangle DBE$.

Hence, $\angle DEB = 140^{\circ}$. Since $\angle AEB = 60^{\circ}$, then $\angle DEA = 80^{\circ}$. Since side $\overline {DE} = \overline {EA}$ in $\triangle DEA$, then $\angle EDA = \angle EAD = 50^{\circ}$.

Now, $\angle EDA$ ($50^{\circ}$) - $\angle EDB$ ($20^{\circ}$) = $\angle CDA$ ($30^{\circ}$).

Answer 2

Hint :

$$sin(140-A)=sin(140)cos(A)-cos(140)sin(A)=ksin(A)$$

with $k=\frac{1}{2sin(70)}$

can be transformed in

$$\frac{sin(140)}{k+cos(140)}=tan(A)$$

Answer 3

By Law of Sines ( using angles in degrees)

$$\dfrac {\sin D }{\sin A}= \dfrac {singlestripledline}{doublestripledline}=\dfrac {singlestripledline }{2\; halfline}=\dfrac {1}{2 \sin 70}$$

$$ 140 = D +A $$

$$ \dfrac{\sin D}{\sin (140-D) } = \dfrac{1/2}{\sin 110} =\dfrac{\sin 30}{\sin 110}$$

Compare arguments,

$ D= 30, 180 + 30, $ the latter angle is discarded.