Question

CDE is equilateral triangle formed on the side CD of a square ABCD. Show that angle EAB= 75°​

Answer 1

Answer:

According to figure,

∠ADC=∠BCD=90

∘

→square

∠CDE=∠DCE=60

∘

→equilateral_triangle

∴∠ADE=∠BCE=150

∘

InΔADEandΔBCE

AD=BC

DE=CE

∠ADE=∠BCE

ΔADE≅ΔBCE

Answer 2

$\large\bold{\underline{\underline{\red{✎Ｇiven:-}}}}$

• ABCD is a Square .
• ∆CDE is a Equalateral Triangle .

To Prove That :-

• EAB = 75°

$\large\bold{\underline{\underline{\red{❒ \: ꜱᴏʟᴜᴛɪᴏɴꜱ:-}}}}$

In AOE ,

∠3 = ∠4 [AO = DE ]

So,∠1 + ∠2 +∠3+∠4 =180°

⟹60°+90°+∠4+∠4 = 180°

⟹2∠4=30°

⟹∠4 = 15°

Now ∠4+∠5 = 90° [ABCD is a Square ]

⟹15°+∠5 = 90°

⟹∠5 = 75°

⟹∠EAB = 75°.

*Figure had Provided check out the

Attachment .

Happy Learning !!