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In quadrilateral $ABCD$, $AB=BC=CD$. The value of $∠BAC$ is $40º$ and the value of $∠CAD$ is $30º$ what is the value of $∠ADC$?

Arnaldo
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Rangan Aryan
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2 Answers2

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For the triangle $ABC$ we have $\angle BCA=\angle BAC =40º$

Sine rule for the triangle $ABC$

$$\frac{AC}{\sin 100º}=\frac{BC}{\sin 40º}$$

Sine rule for the triange $ACD$

$$\frac{AC}{\sin x}=\frac{DC}{\sin 30º}$$

But $BC=DC$ so

$$\frac{\sin x}{\sin 30º}=\frac{\sin 100º}{\sin 40º}$$

Furthermore

$$\sin 100º=\sin 80º=2\sin 40º \cos 40º$$ then

$$\sin x=\cos40º=\sin 50º =\sin 130º\rightarrow x=50º \quad \text{or} \quad x=130º$$

Arnaldo
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  • This solution is great. But, is there any alternative solution without using trigonometry? – Rangan Aryan Dec 30 '16 at 05:38
  • Yes, there is, @Joffan just did, and probably there is another. One way is see the answer and try to construct a geometric solution. That will help you to get some ideas for others problems. – Arnaldo Dec 30 '16 at 11:11
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First note that $\triangle ABC$ is isosceles so $\angle BCA = 40°$ and $\angle ABC = 100°$

Construct $Q$ as the other point of an equilateral triangle on $CD$, towards $A$. Then since $\angle CAD=30°$, $\angle CQD=60°$ and $Q$ is on the bisector of $CD$, $Q$ is the centre of circle $ACD$.

enter image description here

Thus $|AQ|=|CQ|=|CD|$ and $\triangle ACQ$ is isosceles, sharing $AC$ with $\triangle ABC$ and the edges $AB$, $BC$, $AQ$, $CQ$ are the same length. So $\triangle ABC$ and $\triangle ACQ$ are congruent. Now there are two possible cases:

  • As per the diagram, $\triangle ABC$ and $\triangle ACQ$ are mirror images in the shared edge $AC$. Thus $\angle BCD=40+40+60 = 140°$ and $\angle CDA=360-140-100-70 = \fbox{50°}$

  • $\triangle ABC$ and $\triangle ACQ$ are the same triangle, that is, $Q$ is coincident with $B$, giving $\angle BCD = 60°$ and thus $\angle CDA = \fbox{130°}$

(As another intuition on the two results: Given the measures and angles, you can construct $ABC$ and the ray from $A$ towards $D$; then the "swinging gate" of $CD$ can place $D$ in two places along that ray)

Community
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Joffan
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  • Joffan, that was awesome! But could I please know how you figured out so easily that only two possibilities are here. For example, when I looked at your editing, I redrew the figure, and discovered that the point can also coincide with any other point on the quadrilateral, and also it could have been that Q is anywhere interior to the quadrilateral. But, nextly when I tried each of those possibilities, after some logical arguments, it was proved that those were not possible. Finally, it was just the two possibilities in this discussion so far! Wasn't it tough to figure out!! Give me some tips. – Rangan Aryan Jan 05 '17 at 19:45
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    It's certain that $\triangle ABC$ and $\triangle ACQ$ are congruent, and they share edge $AC$. So either $\triangle ACQ$ is the mirror image in $AC$ (as shown) or it is the same triangle ($Q=B$). – Joffan Jan 06 '17 at 12:48
  • I didn't understand , the ittalic lines in the brackets, as given in your answer that you edited on 6th January – Rangan Aryan Jan 14 '17 at 20:55
  • @RanganAryan all I'm saying there is that you know that $|AB|=|BC|=|CD|$ and you can find $\angle BAD$ and $\angle ABC$ with only a little thought; then you can draw those parts, including the direction (but not the length) of $\small\vec{AD}$, place a compass on $C$ with the known length of $CD$ and find two points on $\small \vec {AD}$ to place $D$. The "swinging gate" is just my way of saying that $CD$ can "swing" between the two positions. So - in summary - it's no surprise that there are two answers. – Joffan Jan 14 '17 at 22:54