Calculate $\angle ADC$ in a Convex Quadrilateral

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SUMMARY

In the convex quadrilateral $ABCD$, the angles are defined as follows: $\angle CBD = 10^{\circ}$, $\angle CAD = 20^{\circ}$, $\angle ABD = 40^{\circ}$, and $\angle BAC = 50^{\circ}$. The correct calculation for $\angle ADC$ is determined to be $100^{\circ}$ after adjusting for the convexity of the quadrilateral. The initial assumption of the angle calculation was incorrect due to the misinterpretation of the quadrilateral's shape.

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In a convex quadrilateral $ABCD$ we are given that $\angle CBD=10^{\circ},\,\angle CAD=20^{\circ},\,\angle ABD=40^{\circ},\, \angle BAC=50^{\circ}$.

Find the angle $\angle ADC$.
 
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Re: Find \$\angle ADC\$

My solution (I just solved it):
Let $B$ and $D$ be points on the $x$-axis and extend both the straight lines $BD$ and $AC$ such that they meet at $P$, where $\angle APB=90^{\circ}$ and $BP=k$.

\begin{tikzpicture}[auto]
\draw[thick, ->] (-1,0) -- (10,0) node[anchor = north west] {x};
\draw[thick, ->] (0,-9) -- (0,1) node[anchor = south east] {y};
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=right:P] (P) at (8,0);
\coordinate[label=left: D] (D) at (5,0);
\coordinate[label=right:C] (C) at (8,-3);
\coordinate[label=right:A] (A) at (8,-8);
\draw (B) to (P);
\draw (B) to (A);
\draw (D) to (A);
\draw (B) to (C);
\draw (A) to (P);
\draw (C) to (D);
\draw [dashed] (D) -- (3.6, 1.4);
\node (1) at (1.5,-0.2) {$10^{\circ}$};
\node (2) at (1.5,-0.9) {$30^{\circ}$};
\node (3) at (7,-6.3) {$30^{\circ}$};
\node (4) at (7.7,-6) {$20^{\circ}$};
\draw (P) rectangle +(-0.5, -0.5);
\end{tikzpicture}

We then have

$C=\left(k, -k\tan 10^{\circ}\right)\\D=\left(\dfrac{k}{2\sin 110^{\circ}\cos 40^{\circ}}, 0\right)=\left(\dfrac{k}{\sin 150^{\circ}+\sin 70^{\circ}}, 0\right)=\left(\dfrac{k}{\frac{1}{2}+\cos 20^{\circ}}, 0\right)$

$\begin{align*}\text{Gradient of } l_{CD}&=\dfrac{-k\tan 10^{\circ}}{k\left(1-\dfrac{1}{\frac{1}{2}+\cos 20^{\circ}}\right)}\\&=\dfrac{-\tan 10^{\circ}\left(\frac{1}{2}+\cos 20^{\circ}\right)}{\cos 20^{\circ}-\dfrac{1}{2}}\\&=\dfrac{-\tan 10^{\circ}(4\cos^2 10^{\circ}-1)}{4\cos^2 10^{\circ}-3}\\&=\dfrac{-\tan 10^{\circ}(4\cos^2 10^{\circ}-(\sin^2 10^{\circ} +\cos^2 10^{\circ}))}{4\cos^2 10^{\circ}-3(\sin^2 10^{\circ} +\cos^2 10^{\circ})}\\&=\dfrac{-\tan 10^{\circ}(3\cos^2 10^{\circ}-\sin^2 10^{\circ})}{\cos^2 10^{\circ}-3\sin^2 10^{\circ}}\\&=\dfrac{-\tan 10^{\circ}(3-\tan^2 10^{\circ})}{1-3\tan^2 10^{\circ}}\\&=-\tan (3\times 10)^{\circ}\\&=\tan 150^{\circ}\end{align*}$

This gives $\angle CDP=180^{\circ}-150^{\circ}=30^{\circ}$.

But $\angle ACP=30^{\circ}+40^{\circ}=70^{\circ}$, we get $\angle ADC=70^{\circ}-30^{\circ}=40^{\circ}$.
 
Re: Find \$\angle ADC\$

anemone said:
My solution (I just solved it):
Let $B$ and $D$ be points on the $x$-axis and extend both the straight lines $BD$ and $AC$ such that they meet at $P$, where $\angle APB=90^{\circ}$ and $BP=k$.

\begin{tikzpicture}[auto]
\draw[thick, ->] (-1,0) -- (10,0) node[anchor = north west] {x};
\draw[thick, ->] (0,-9) -- (0,1) node[anchor = south east] {y};
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=right:P] (P) at (8,0);
\coordinate[label=left: D] (D) at (5,0);
\coordinate[label=right:C] (C) at (8,-3);
\coordinate[label=right:A] (A) at (8,-8);
\draw (B) to (P);
\draw (B) to (A);
\draw (D) to (A);
\draw (B) to (C);
\draw (A) to (P);
\draw (C) to (D);
\draw [dashed] (D) -- (3.6, 1.4);
\node (1) at (1.5,-0.2) {$10^{\circ}$};
\node (2) at (1.5,-0.9) {$30^{\circ}$};
\node (3) at (7,-6.3) {$30^{\circ}$};
\node (4) at (7.7,-6) {$20^{\circ}$};
\draw (P) rectangle +(-0.5, -0.5);
\end{tikzpicture}

We then have

$C=\left(k, -k\tan 10^{\circ}\right)\\D=\left(\dfrac{k}{2\sin 110^{\circ}\cos 40^{\circ}}, 0\right)=\left(\dfrac{k}{\sin 150^{\circ}+\sin 70^{\circ}}, 0\right)=\left(\dfrac{k}{\frac{1}{2}+\cos 20^{\circ}}, 0\right)$

$\begin{align*}\text{Gradient of } l_{CD}&=\dfrac{-k\tan 10^{\circ}}{k\left(1-\dfrac{1}{\frac{1}{2}+\cos 20^{\circ}}\right)}\\&=\dfrac{-\tan 10^{\circ}\left(\frac{1}{2}+\cos 20^{\circ}\right)}{\cos 20^{\circ}-\dfrac{1}{2}}\\&=\dfrac{-\tan 10^{\circ}(4\cos^2 10^{\circ}-1)}{4\cos^2 10^{\circ}-3}\\&=\dfrac{-\tan 10^{\circ}(4\cos^2 10^{\circ}-(\sin^2 10^{\circ} +\cos^2 10^{\circ}))}{4\cos^2 10^{\circ}-3(\sin^2 10^{\circ} +\cos^2 10^{\circ})}\\&=\dfrac{-\tan 10^{\circ}(3\cos^2 10^{\circ}-\sin^2 10^{\circ})}{\cos^2 10^{\circ}-3\sin^2 10^{\circ}}\\&=\dfrac{-\tan 10^{\circ}(3-\tan^2 10^{\circ})}{1-3\tan^2 10^{\circ}}\\&=-\tan (3\times 10)^{\circ}\\&=\tan 150^{\circ}\end{align*}$

This gives $\angle CDP=180^{\circ}-150^{\circ}=30^{\circ}$.

But $\angle ACP=30^{\circ}+40^{\circ}=70^{\circ}$, we get $\angle ADC=70^{\circ}-30^{\circ}=40^{\circ}$.
How do we know that AC is perpendicular to BD
 
Re: Find \$\angle ADC\$

kaliprasad said:
How do we know that AC is perpendicular to BD

Hi kaliprasad!

Notice that $\angle A=50^{\circ}$ and $\angle B=40^{\circ}$, if we extend the lines $AC$ and $BD$, to form a triangle, we see that we will get a right-angled triangle.
 
Re: Find \$\angle ADC\$

anemone said:
My solution (I just solved it):
Let $B$ and $D$ be points on the $x$-axis and extend both the straight lines $BD$ and $AC$ such that they meet at $P$, where $\angle APB=90^{\circ}$ and $BP=k$.

\begin{tikzpicture}[auto]
\draw[thick, ->] (-1,0) -- (10,0) node[anchor = north west] {x};
\draw[thick, ->] (0,-9) -- (0,1) node[anchor = south east] {y};
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=right:P] (P) at (8,0);
\coordinate[label=left: D] (D) at (5,0);
\coordinate[label=right:C] (C) at (8,-3);
\coordinate[label=right:A] (A) at (8,-8);
\draw (B) to (P);
\draw (B) to (A);
\draw (D) to (A);
\draw (B) to (C);
\draw (A) to (P);
\draw (C) to (D);
\draw [dashed] (D) -- (3.6, 1.4);
\node (1) at (1.5,-0.2) {$10^{\circ}$};
\node (2) at (1.5,-0.9) {$30^{\circ}$};
\node (3) at (7,-6.3) {$30^{\circ}$};
\node (4) at (7.7,-6) {$20^{\circ}$};
\draw (P) rectangle +(-0.5, -0.5);
\end{tikzpicture}

We then have

$C=\left(k, -k\tan 10^{\circ}\right)\\D=\left(\dfrac{k}{2\sin 110^{\circ}\cos 40^{\circ}}, 0\right)=\left(\dfrac{k}{\sin 150^{\circ}+\sin 70^{\circ}}, 0\right)=\left(\dfrac{k}{\frac{1}{2}+\cos 20^{\circ}}, 0\right)$

$\begin{align*}\text{Gradient of } l_{CD}&=\dfrac{-k\tan 10^{\circ}}{k\left(1-\dfrac{1}{\frac{1}{2}+\cos 20^{\circ}}\right)}\\&=\dfrac{-\tan 10^{\circ}\left(\frac{1}{2}+\cos 20^{\circ}\right)}{\cos 20^{\circ}-\dfrac{1}{2}}\\&=\dfrac{-\tan 10^{\circ}(4\cos^2 10^{\circ}-1)}{4\cos^2 10^{\circ}-3}\\&=\dfrac{-\tan 10^{\circ}(4\cos^2 10^{\circ}-(\sin^2 10^{\circ} +\cos^2 10^{\circ}))}{4\cos^2 10^{\circ}-3(\sin^2 10^{\circ} +\cos^2 10^{\circ})}\\&=\dfrac{-\tan 10^{\circ}(3\cos^2 10^{\circ}-\sin^2 10^{\circ})}{\cos^2 10^{\circ}-3\sin^2 10^{\circ}}\\&=\dfrac{-\tan 10^{\circ}(3-\tan^2 10^{\circ})}{1-3\tan^2 10^{\circ}}\\&=-\tan (3\times 10)^{\circ}\\&=\tan 150^{\circ}\end{align*}$

This gives $\angle CDP=180^{\circ}-150^{\circ}=30^{\circ}$.

But $\angle ACP=30^{\circ}+40^{\circ}=70^{\circ}$, we get $\angle ADC=70^{\circ}-30^{\circ}=40^{\circ}$.
[sp]Very nice solution, anemone! My only criticism is that the quadrilateral $ABCD$ is supposed to be convex. If the vertices $A,B,C,D$ are to be taken in that order then your quadrilateral is not convex. I think that the diagram should look more like this:
[TIKZ] [scale=0.75]
\coordinate[label=left:A] (A) at (-10,0);
\coordinate[label=right:B] (B) at (0,12);
\coordinate[label=right:C] (C) at (2,0);
\coordinate[label=below right: D] (D) at (0,-3.5);
\draw (A) -- (B) -- (C) -- (D) -- (A) -- (C);
\draw (B) -- (D);
\node at (0.7,10.5) {$\leftarrow 10^{\circ}$};
\node at (-8,-0.3) {$20^{\circ}$};
\node at (-0.5,10.5) {$40^{\circ}$};
\node at (-8.5,0.5) {$50^{\circ}$};
\draw (0,0) rectangle +(-0.5, 0.5);
[/TIKZ]
Your trigonometric argument then works almost exactly as before, but the final result is that $\angle ADC$ is then $70^{\circ} + 30^{\circ}$ instead of $70^{\circ}-30^{\circ}$. So I think that the answer should be $100^\circ$.[/sp]
 
Re: Find \$\angle ADC\$

I initially struggled as I didn't know for sure if the vertices $A,B,C,D$ are to be taken in that order or not...after I saw there was no reply to this challenge, I decided to go with my hunch that it is okay they are not ordered in alphabetical manner. Now, the more I read it, the more I think you are right, so, thanks so much to Opalg for pointing it out...

Here is the amended solution (with much more neater trigonometric argument(Sun)):
[TIKZ] [scale=0.75]
\coordinate[label=left:A] (A) at (-10,0);
\coordinate[label=right:B] (B) at (0,12);
\coordinate[label=right:C] (C) at (2,0);
\coordinate[label=below right: D] (D) at (0,-3.5);
\coordinate[label=below right:P] (P) at (0,0);
\draw (A) -- (B) -- (C) -- (D) -- (A) -- (C);
\draw (B) -- (D);
\node at (0.7,10.5) {$\leftarrow 10^{\circ}$};
\node at (-8,-0.3) {$20^{\circ}$};
\node at (-0.5,10.5) {$40^{\circ}$};
\node at (-8.5,0.5) {$50^{\circ}$};
\draw (0,0) rectangle +(-0.5, 0.5);
[/TIKZ]Let $A$ be the origin point in the Cartesian plan, $AC$ and $BD$ meet at $P$ and $AP=k$.

We then have

$D=\left(k, -k\tan 20^{\circ}\right)\\C=\left(k\left(1+\dfrac{\tan 10^{\circ}(1-\tan^2 20^{\circ})}{2\tan 20^{\circ}}\right), 0\right)=\left(k\left(\dfrac{2\tan 20^{\circ}+\tan 10^{\circ}(1-\tan^2 20^{\circ})}{2\tan 20^{\circ}}\right), 0\right)$

$\begin{align*}\text{Gradient of } l_{CD}&=\dfrac{-k\tan 20^{\circ}}{k\left(1-\dfrac{2\tan 20^{\circ}+\tan 10^{\circ}(1-\tan^2 20^{\circ})}{2\tan 20^{\circ}}\right)}\\&=\dfrac{-\tan 20^{\circ}\left(2\tan 20^{\circ}\right)}{-\tan 10^{\circ}(1-\tan^2 20^{\circ})}\\&=\tan20^{\circ}\tan40^{\circ} \tan80^{\circ}\\&=\tan (3\times 20)^{\circ}\\&=\tan60^{\circ}\end{align*}$

This means $\angle PCD=60^{\circ}$ and therefore $\angle PDC=30^{\circ}$.

This implies $\angle ADC=70^{\circ}+30^{\circ}=100^{\circ}$.
 

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