Calculate $\angle ADC$ in a Convex Quadrilateral

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In summary, In a convex quadrilateral $ABCD$ with $\angle CBD=10^{\circ},\,\angle CAD=20^{\circ},\,\angle ABD=40^{\circ},\, \angle BAC=50^{\circ}$, the angle $\angle ADC$ is $100^\circ$.
  • #1
anemone
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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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  • #2
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 $\angle ADC=70^{\circ}-30^{\circ}=40^{\circ}$.
 
  • #3
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
 
  • #4
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.
 
  • #5
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]
 
  • #6
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 $\angle ADC=70^{\circ}+30^{\circ}=100^{\circ}$.
 

FAQ: Calculate $\angle ADC$ in a Convex Quadrilateral

What is a convex quadrilateral?

A convex quadrilateral is a four-sided polygon with all interior angles less than 180 degrees. This means that all the vertices of the quadrilateral point outwards, creating a "bulging" shape.

How do you calculate the measure of angle ADC in a convex quadrilateral?

To calculate the measure of angle ADC, you can use the formula: angle ADC = 180 - (angle ABC + angle CDA). This formula is based on the fact that the sum of all angles in a convex quadrilateral is always 360 degrees.

Can you provide an example of calculating angle ADC in a convex quadrilateral?

Sure, let's say we have a convex quadrilateral with angles ABC measuring 60 degrees and CDA measuring 120 degrees. Using the formula mentioned above, we can calculate angle ADC as follows: angle ADC = 180 - (60 + 120) = 180 - 180 = 0 degrees. This means that angle ADC is a straight angle, measuring 180 degrees.

What if the convex quadrilateral is not given in a diagram?

In that case, you will need to know the measures of at least three angles in the quadrilateral in order to calculate the measure of angle ADC. You can use the formula mentioned in question 2 and solve for angle ADC using the given angle measures.

Is there any other method to calculate angle ADC in a convex quadrilateral?

Yes, there is another method known as the "exterior angle sum property". According to this property, the measure of an exterior angle of a convex quadrilateral is equal to the sum of the two opposite interior angles. So, to calculate angle ADC, you can subtract angle ABC from 180 degrees and then subtract angle CDA from the result. This will give you the measure of angle ADC.

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