In Fig. 104, AB || CO, ∠OAB = 150° and ∠OCO = 120°. Then, ∠AOC =
(a) 80°
(b) 90°
(c) 70°
(d) 100°

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Solution

Construction: Draw a line OE from the point O parallel to AB and CD

Since, AB || OE
∴ ∠BAO + ∠AOE = 180^∘ [Angles on the same side of a transversal line are supplementary]
⇒ 150^∘ + ∠AOE = 180^∘
⇒ ∠AOE = 30^∘
Again, CD || OE
∴ ∠DCO + ∠COE = 180^∘ [Angles on the same side of a transversal line are supplementary]
⇒ 120^∘ + ∠COE = 180^∘
⇒ ∠COE = 60^∘
Now, ∠AOC = ∠AOE + ∠COE
= 30^∘ + 60^∘
= 90^∘
Hence, the correct answer is option (b).

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In the given figure, AB || CD. ∠OAB = 150° and ∠OCD = 120°.
Then ∠AOC = ?

(a) 80°
(b) 90°
(c) 70°
(d) 100°

Q. In the given figure, AB || CD. If ∠OAB = 124° and ∠OCD = 136°, then ∠AOC = ?
(a) 80°
(b) 90°
(c) 100°
(d) 110°

Q. In the given figure, AB || CD. If ∠AOC = 30° and ∠OAB = 100°, then ∠OCD = ?
(a) 130°
(b) 150°
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In the given figure, AB||CD. If $∠ A O C = 30^{\circ}$ and $∠ O A B = 100^{\circ}$ then $∠ O C D = ?$

$(a) 130^{\circ}$

$(b) 150^{\circ}$

$(c) 80^{\circ}$

$(d) 100^{\circ}$

The median for the data is

$\begin{matrix} I Q & 60 - 70 & 70 - 80 & 80 - 90 & 90 - 100 & 100 - 110 & 110 - 120 & 120 - 130 N o . o f p u p i l s & 2 & 3 & 5 & 16 & 14 & 13 & 7 \end{matrix}$

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In Fig. 104, AB || CO, ∠OAB = 150° and ∠OCO = 120°. Then, ∠AOC = (a) 80° (b) 90° (c) 70° (d) 100°

In Fig. 104, AB || CO, ∠OAB = 150° and ∠OCO = 120°. Then, ∠AOC =
(a) 80°
(b) 90°
(c) 70°
(d) 100°