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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 $\color{red}{30^\circ}$.]

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  • $\begingroup$ Isn't \triangle ACD : $\triangle ACD$ a better symbol than '\overset\Delta`...? $\endgroup$ – CiaPan Nov 3 '15 at 12:04
  • $\begingroup$ @Matthias, yes, forgot that. I will edit the question now. Thank you. $\endgroup$ – Alistair Nov 3 '15 at 12:10
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enter image description hereObviously, $\angle BAC = \angle BCA = 20^{\circ}$, 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, $\angle EDA$ ($50^{\circ}$) - $\angle EDB$ ($20^{\circ}$) = $\angle CDA$ ($30^{\circ}$).

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    $\begingroup$ Once we know that $EA=EB=ED$ we can just observe that $ADB$ and $AEB$ are inscribed angle and central angle respectively subtending arc $AB$ in the circle centered at $E$ and radius $EA$ which implies that $\angle ADB = \frac 12 \angle AEB = 30^\circ$. $\endgroup$ – timon92 Nov 3 '15 at 22:19
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Hint :

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

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

can be transformed in

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

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  • $\begingroup$ You can replace $A$ by $D$ and $k=\frac{1}{2sin(70)}$ by $k=2sin(70)$ to get $tan(D)$. $\endgroup$ – Peter Nov 3 '15 at 13:07
  • $\begingroup$ To show, that the result for $D$ is exact 30 (not just near 30), you have to show $\frac{sin(40)}{2sin(70)-cos(40)}=\frac{1}{3}\sqrt{3}$ $\endgroup$ – Peter Nov 3 '15 at 13:09
  • $\begingroup$ Ah, this was a nice and obvious trick! It's very easy to show the last equation. Thanks. $\endgroup$ – Alistair Nov 3 '15 at 13:53
  • $\begingroup$ @Peter I find this hint obscure to me ,i mean how did you know that $\sin (140 - A) =k \sin(A) $ would work here, since this relation it's not always true ? sorry if i am asking on another question,but that hint intrigued me.I hope you will answer me.Thanks in advance. $\endgroup$ – Nameless Nov 3 '15 at 14:59
  • $\begingroup$ We have $\frac{sin(140-A)}{sin(A)}=\frac{sin(D)}{sin(A)}=\frac{1}{2sin(70)}\ =:\ k$. The other equation comes from $sin(x-y)=sin(x)cos(y)-cos(x)sin(y)$ $\endgroup$ – Peter Nov 3 '15 at 18:30
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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.

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    $\begingroup$ I already had this with the law of sines. But i couldn't find anything using trigonometry, without exploiting the known result. $\endgroup$ – Alistair Nov 3 '15 at 12:34
  • $\begingroup$ @Peter Thanks, corrected. $\endgroup$ – Narasimham Nov 3 '15 at 12:38
  • $\begingroup$ @Alistair Trig is geometry applied to triangles, for quicker computation. Whatever you do, it will come back to this configuration based computations. $\endgroup$ – Narasimham Nov 3 '15 at 12:42
  • $\begingroup$ @Narasimham, i know. I already had this configuration, but i couldn't find the answer from there. $\endgroup$ – Alistair Nov 3 '15 at 12:46
  • $\begingroup$ $A$ is the angle in the triangle ABD, not ACD. $\endgroup$ – Peter Nov 3 '15 at 12:58

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