Question

pls don't spam and I need answer on copy​

Answer 1

Answer:

(I)ΔABC≅ΔBAD (SSS)

∵BC=AD

 AB is common

 AC=BD (diagonal are equal)

∴∠BAC=∠ABD=35°

X=90°-∠ABD

⇒X=15°

In ΔABC

∠ACB=180°-∠CBA-∠BAC

⇒∠ACB=15°

Y=X+∠ACB (EXTERNAL ANGLE PROPERTY)

⇒Y=30°

(II) X+Y=90°

∠ABD=∠ACD=Y

∠DBC=90°-Y

Again Applying external angle property in ΔOBC

∴∠ABO=∠OBC+∠OCB

⇒110°=90-Y+X

⇒20°=X-Y

X=55°

Y=35°

correct answer hai ya nai friend