Question

in the figure given below along the side ABCD is a square and triangle DEC is an equilateral triangle .prove that angle DAE is equal to 15 degree . i will mark this answer as a brainalist
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Answer 1

Answer:

The ∠DAE is equal to 15°.

Proved.

Step-by-step explanation:

Given -

ABCD is a square and EDC is an equilateral triangle.

To Prove -

The ∠DAE is equals to 15°.

Proof -

Let's Start from,

⇒ AD = BD   [ Side of same square ]

⇒ ∠ADE = ∠BCE    [ Each angle = 150° ]

⇒ DE = BE    [ Side of same equilateral triangle ]

As we know,

The SAS Rule

Statement -

If the Two sides and the angle of one triangle are congruent to two sides and the angle of another triangle, then these two triangles are congruent.

So,

It implies,

ΔADE ≅ ΔBCE.

∴  AE = BE.

Now,

In ΔADE

⇒ AD = DE

⇒ ∠DAE = ∠DEC.

In ΔADE,

⇒∠DAE +∠DEC + ∠ADE = 180°.

{ The sum of all angles of triangle is 180° }

$\sf\\ \\\implies 2\angle DAE+150^{\circ}=180^{\circ}\\ \\ \implies 2\angle DAE=180^{\circ}-150^{\circ}\\ \\\implies 2\angle DAE=30^{\circ}\\ \\ \implies \angle DAE = \sqrt{30}\\ \\ \implies \angle DAE=15^{\circ}$

__________{ Proved }

Answer 2

Given :-

ABCD is a square

DEC is an equilateral triangle

To Prove :-

∠DAE = 15°

Solution :-  

AD = BD   [ Side of square are equal]

∠ADE = ∠BCE    [ Each angle 150° ]

DE = BE    [ sides of equilateral triangle are equal ]

ΔADE ≅ ΔBCE. [ By SAS congruency rule ]

AE = BE [ C.P.C.T ]

In ΔADE

AD = DE

∠DAE = ∠DEA [ Angle opposite to equal side ]

In ΔADE,

⇒∠DAE +∠DEC + ∠ADE = 180° [ ASP of triangle ]

2∠DAE+150∘=180∘

2∠DAE=180∘−150∘

2∠DAE=30∘

∠DAE=30

∠DAE=15∘

Hence proved !