Housing an Angle

Geometry Level 1

In the diagram above, A B C D ABCD is a square and C D E CDE is an equilaterial triangle.

What is ∠ D A E ? \angle DAE?

1 0 ∘ 10 ^ \circ 1 5 ∘ 15 ^ \circ 2 0 ∘ 20 ^ \circ 3 0 ∘ 30 ^ \circ

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Chung Kevin by Chung Kevin , 45, Australia

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4 solutions

Marta Reece
Apr 10, 2017

In the isosceles △ A D E \triangle ADE from the sum of angles ∠ D A E = 18 0 ∘ − ( 9 0 ∘ + 6 0 ∘ ) 2 = 1 5 ∘ \angle DAE=\frac{180^\circ-(90^\circ+60^\circ)}{2}=15^\circ

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AD is equal to DE, so it is an isosceles triangle. Knowing that the angle ADE is 150 (90+60) [60 because DEC is a equilaterial triangle], so the sum of DAE and DEA angles must be 30. Once they are the same, 30/2 = 15

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Might be a stupid question, but how do you know AD = DE?

Karun Mukesh Jhangiani - 4 years, 1 month ago

An equilateral triangle is supposed to have congruent sides So AB = BC = CD = AD = DE =EC

Ajay Sambhriya - 3 years, 10 months ago
Adrian Ciurea
May 3, 2017

3 Helpful 0 Interesting 0 Brilliant 0 Confused

explaining what he basically did, by simetry he saw that the left angle alpha at (A) its equal to left angle on (E), then he cut the square to make it into 2 rectangular triangles ( its not really necessary ) but now he can clearly see that the angle down on point D is 90 ° (like i said its not necessary because u can clearly see that) since triangle DEC is isosceles angle (D) (E) and (C) are all equals and 60 ° now he know that angle D = 90 ° + 60 °, also that A and E (left angles) are equals so he had the relation 180 ° - 150 ° = 2a (I explaned that because i can't read it but i can read some things and my solution was very similiar to his solution.

aaaaaaa aaaaaa - 4 years, 1 month ago

Consider △ A D E \triangle ADE . Since A D = D E AD=DE , ∠ D A E = ∠ D E A \angle DAE = \angle DEA .

∠ E D A = ∠ E D C + ∠ C D A = 60 + 90 = 15 0 ∘ \angle EDA=\angle EDC + \angle CDA= 60 + 90 = 150^\circ

Since the sum of the interior angles of a triangle is 18 0 ∘ 180^\circ , we have

∠ D A E = ∠ D E A = 180 − 150 2 = 30 2 = 1 5 ∘ \angle DAE = \angle DEA = \dfrac{180-150}{2}=\dfrac{30}{2}=\boxed{15^\circ}

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