4

enter image description here

Can you help me to determine the angle marked with a question mark?

$\overline {AB}$ and $\overline {DE}$ are parallel

suomynonA
  • 6,895
romanemul
  • 63

4 Answers4

2

HINT: Draw a line through $C$ that is parallel to $AB$ and $DE$. This splits the angle in question into two parts. Can you find the measures of the two parts using alternate interior angles?

scott
  • 1,607
2

enter image description here

Divide $\angle C$ into angles $C_1$ and $C_2$ as shown.

  • Since $\overline {AB}$ and $\overline {CF}$ are parallel, then $\angle C_2=40^\circ$ by alternate interior angles.
  • You can see that the supplement of $\angle D$ is $60^\circ$. Since $\overline {DE}$ and $\overline {CF}$ are parallel, then $\angle C_1=60^\circ$ by alternate interior angles.

Add up $\angle C_1$ and $\angle C_2$ and you get $40^\circ+60^\circ=\color{red}{100^\circ}$

suomynonA
  • 6,895
1

Assuming lines $AB$ and $DE$ are parallel, you can draw a line perpendicular to the line $DE$ (let's call this point $M$) to the point $B$ to obtain the quadrilateral $BCDM$.

Now, use the fact that the sum of the interior angles in a quadrilateral is $360^{\circ}$.

From this, you can deduce that angles $\angle{CBM}=90^{\circ}-40^{\circ}=50^{\circ}$ and $\angle{BMD}=90^{\circ}$.

Thus, the angle of the question mark is:

$\angle{DCB}=360^{\circ}-50^{\circ}-90^{\circ}-120^{\circ}=100^{\circ}$

projectilemotion
  • 15,428
1

Draw a line perpendicular to $AB$ through $C$.

Then $\angle A'CB=180^\circ-90^\circ-40^\circ=90^\circ-40^\circ=50^\circ$.

Do the same for $DE$ to get $\angle D'CE=90^\circ-60^\circ=30^\circ$.

Now $\angle BCD=180^\circ-(50^\circ+30^\circ)=100^\circ$.

This, conveniently, is $\angle ABC+(180^\circ-\angle EDC)$.

JMP
  • 21,771