Question

in∆XYZ seg xy≈segxz,YQ and seg.zp are the angle Bisector of angle and y angle z respectively angle x 28° then find angle YOZ​

Answer 1

$/bold{Solution}$

$\bold{in \: ∆xyz} \\ \tt \bold{given} \\ \longmapsto \tt \: < xyz = 54degree \\ \longmapsto \tt < x = 62degree \\ \bold{to \: find} \\ \longmapsto \tt < ozy \\ \longmapsto \tt < yoz \\ \\ \bold{solution : :} \\ \\ \longmapsto \tt < x + < y + < z = 180degree(a.s.p) \\ \longmapsto \tt 62degree + 54degree + < z = 180degree \\ \longmapsto \tt < z = 180degree - 116degree \\ \longmapsto \: < z = 64degree \\ \bold{so} \\ \longmapsto \tt < ozy = \frac{54}{2} = 27degree \\ \bold{similiary} \\ \longmapsto \: < ozy = \frac{64}{2} = 32degree \\ \bold{now} \\ \bold{ \:∆yoz} \\ \longmapsto \tt < yoz + < oyz + < ozy = 180degree(a.s.p) \\ \longmapsto \tt < yoz + 27degree + 32degree = 180degree \\ \longmapsto \tt < yoz + 59degree = 180degree \\ \longmapsto \tt < yoz = 180degree - 59degree \\ \longmapsto < yoz = 121degree$

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Answer 2

Answer:

in∆xyz

given

⟼<xyz=54degree

⟼<x=62degree

tofind

⟼<ozy

⟼<yoz

solution::

⟼<x+<y+<z=180degree(a.s.p)

⟼62degree+54degree+<z=180degree

⟼<z=180degree−116degree

⟼<z=64degree

so

⟼<ozy=

2

54

=27degree

similiary

⟼<ozy=

2

64

=32degree

now

∆yoz

⟼<yoz+<oyz+<ozy=180degree(a.s.p)

⟼<yoz+27degree+32degree=180degree

⟼<yoz+59degree=180degree

⟼<yoz=180degree−59degree

⟼<yoz=121degree