In the figure PA and PB are tangents to a circle with centre O if angle AOB=120degrees find the angle OPA
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Answered by
77
angle AOB =120°
angle PAO=angle PBO=90°
(Tangent theorem )
AOBP is a quadrilateral
the sum of all angles of an quadrilateral is 360°.
AOB+OBP+BPA+PAO=360°
120°+90°+90°+BPA=360°
BPA=360°-300°
BPA=60°.
OPB=OPA....1
BPA=BPO+APO
60°=2OPA From 1
OPA=30°
hope this helps you:-)))))
don't forget to do it brainleist.
angle PAO=angle PBO=90°
(Tangent theorem )
AOBP is a quadrilateral
the sum of all angles of an quadrilateral is 360°.
AOB+OBP+BPA+PAO=360°
120°+90°+90°+BPA=360°
BPA=360°-300°
BPA=60°.
OPB=OPA....1
BPA=BPO+APO
60°=2OPA From 1
OPA=30°
hope this helps you:-)))))
don't forget to do it brainleist.
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chhajedchirag1p32bb2:
I have forgotten one step triangle PAO congruent to triangle PBO by hypotenuse side test. therefore angle APO=BPO......c a.c.t
Answered by
28
Answer=^opa=30°
Step-by-step explanation:
Given-oap=obp=90°
construction-join op
Proof-in triangle oap
^oap+^opa+^aop=180°
90+opa+60°=180°
opa=180-150
^opa=30°
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