Question

In the figure shown, ABCD is a square and triangle CDE is equilateral. What is the degree measure of angle CBE?

Answer 1

15 degrees

Explanation:

Since triangle CDE is a equilateral triangle then all the sides and angles are congruent DC=CE=ED, ∠DEC=∠ECD=∠CDE=60 degrees

since line CD is also a side of the square and all sides of a square are congruent therefore DC=CB=BA=AD

therefore side BC is equal to side CE which means CBE is an isosceles triangle then angles ∠BEC and ∠CBE are congruent

Then in order to figure out angle ∠CBE we need to find ∠ECB

and ∠ECB=∠DCE+∠DCB

∠DCE= 60 degrees because it's a equilateral triangle

∠DCB=90 degrees because it's a square

therefore ∠ECB=150 degrees

all angles of a triangle = 180 degrees

∠BEC + ∠CBE + ∠ECB = 180

since ∠BEC is congruent to ∠CBE i can substitute ∠BEC for ∠CBE

2(∠CBE) + 150 =180

2(∠CBE)=30

∠CBE=15

Answer 2

angle CBE=15 degrees

Explanation:

angle ECD is 60 degrees and EC=CD because equilateral triangle

angle DCB is 90 degrees and CB=CD because square

angle ECD +angle DCB= angle ECB

60+90=150=angle ECB

since triangle ECB has legs EC and CB equal in length, then it is an isosceles triangle.

so angles CBB and CBE are equal

angle CBE=(180-150)/2

angle CBE=15 degrees