In the adjoining figure, $A B C D$ is a square and $△ E D C$ is an equilateral triangle. Prove that
$∠ D A E = 15^{o}$ .

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Solution

$∠ D A E + ∠ A D E + ∠ D E A = 180^{o}$
$150^{o} + ∠ A D E + ∠ D E A = 180^{o}$
$2 ∠ D A E = 180^{o} - 150^{o}$
$2 ∠ D A E = 30^{o}$
$∠ D A E = 15^{o}$

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Similar questions

In the adjoining figure, $A B C D$ is a square and $Δ E D C$ is an equilateral triangle. Prove that (i) $A E = B E,$ (ii) $∠ D A E = 15^{\circ}$ .

In the adjoining figure, $A B C D$ is a square and $△ E D C$ is an equilateral triangle. Prove that
$A E = B E$ .

A B C D

△ A P B

△ A P D ≅ △ B P C

A B C D

E D C

∠ D A B = 48^{o}

∠ B E C

∠ D E B

∠ B F C

∠ D A E

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