In the adjoining figure- it is given that displaystyle ABparallel CD- angle BAO-108-circ- and angle OCD - 120-circ- then angle AOC-displaystyle 120-circ-displaystyle 132-circ-displaystyle 150-circ-displaystyle 72-circ-

Question

Toppr Admin · Accepted Answer

Through O draw a line EOF parallel to AB
now EF || AB and CD||AB
$∵$ AB || EF and AO is a transversal
we have
$∠ A O F + ∠ O A B = 180^{\circ}$
$∠ A O F + 108^{\circ} = 180^{\circ}$
$∠ A O F = 72^{\circ}$
$∵$ EF|| CD and OC is transversal
so
$∠ C O F + ∠ O C D = 180^{\circ}$
$∠ C O F + 120^{\circ} = 180^{\circ}$
$∠ C O F = 60^{\circ}$
$∴ ∠ A O C = ∠ A O F + ∠ C O F$
$= 72^{\circ} + 60^{\circ} = 132^{\circ}$

In the adjoining figure, it is given that AB∥CD,∠BAO=108∘ and ∠OCD=120∘, then ∠AOC=_______.120∘132∘150∘72∘

Through O draw a line EOF parallel to ABnow EF || AB and CD||AB∵ AB || EF and AO is a transversal we have∠AOF+∠OAB=180∘∠AOF+108∘=180∘∠AOF=72∘∵ EF|| CD and OC is transversal so∠COF+∠OCD=180∘∠COF+120∘=180∘∠COF=60∘∴∠AOC=∠AOF+∠COF=72∘+60∘=132∘

In the adjoining figure, it is given that $A B ∥ C D, ∠ B A O = 108^{\circ}$ and $∠ O C D = 120^{\circ},$ then $∠ A O C =$ _______.

$120^{\circ}$
$132^{\circ}$
$150^{\circ}$
$72^{\circ}$