Question

In the figure, $\angle A,$ $\angle B$ and $\angle C$ are right angles. If $\angle AEB = 40^\circ$ and $\angle BED = \angle BDE,$ then $\angle CDE =$

$75^\circ$

$80^\circ$

$85^\circ$

$90^\circ$

$95^\circ$

Answer: E

Concepts:angle chasing angle sum

Difficulty rating: 1150

Solution:

In triangle $BDE,$ the angles at $E$ and $D$ are equal and $\angle B = 90^\circ,$ so $\angle BED = \angle BDE = 45^\circ.$

Then $\angle AED = \angle AEB + \angle BED$ $= 40^\circ + 45^\circ$ $= 85^\circ.$ In quadrilateral $AEDC,$ the angles at $A$ and $C$ are $90^\circ,$ so

$\begin{aligned} \angle CDE &= 360^\circ - 90^\circ - 90^\circ \\ &\quad {}- 85^\circ \\ &= 95^\circ. \end{aligned}$

Thus, the correct answer is E .

1995 AMC 8 Problem 13

Answer: E

Difficulty rating: 1150

Solution:

Problem 13 in Other Years