Question

4th no. Please Solve fast.....

Answer 1

hope this solution help you

Answer 2

Ray BO is the bisector angel CBE

angel CBO=angel EBO=1\2 angel CBE

angel CBO=1\2(180°-y)

angel CBO=90-y\2

ray CO is bisector of angel BCD

angel BCO=angel OCD=1\2 angel BCD

angel BCO=1\2 (180-°z)

angel BCO=90°-z\2

In ∆BCO

angel BCO + angel BOC + CBO =180°

90°-z\2 + angel BOC + 90 -y\2=180°

angel BOC=z\2+y\2

angel BOC =y+2\2———(eq.1)

In ∆ABC

angel A + angel B + angel C=180°

X+Y+Z=180°

Y+Z=180°-x. ———(eq.2)

angel BOC=18-x\2

angel=90°-x\2

angel=90°- angel BAC\2

$hope \: it \: is \: helpful$

$please \: mark \: as \: brainleast$