In the adjoining figure, it is given that AB∥CD,∠BAO=108∘ and ∠OCD=120∘, then ∠AOC=_______.
120∘
132∘
150∘
72∘
A
120∘
B
150∘
C
72∘
D
132∘
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Solution
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Through O draw a line EOF parallel to AB now 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∘
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