Question

In the given figure,$\angle OAB={75}^{\circ },\angle OBA={55}^{\circ }$ and $\angle OCD={100}^{\circ }$. Then, $\angle ODC=?$


$\left(a\right){20}^{\circ }$

$\left(b\right){25}^{\circ }$

$\left(c\right){30}^{\circ }$

$\left(d\right){35}^{\circ }$

Answer 1

(c) 30°

In ∆OAB, we have:

∠OAB+∠OBA+∠AOB=180° [ Sum of the angles of a triangle]

⇒75°+55°+∠AOB=180°

⇒∠AOB=50°

∴∠COD=∠AOB=50° Vertically-Opposite Angles

In ∆OCD, we have:

∠COD+∠OCD+∠ODC=180° [Sum of the angles of a triangle]

⇒50°+100°+∠ODC=180°

⇒∠ODC=30°