Question

5. In the given figure, O is the centre of the
circle. If angleACB = 50°, find angleOAB.​

Answer 1

Answer:

130

Step-by-step explanation:

OAB + ACB = 180

OAB + 50 = 180

OAB = 180-50

OAB = 130

Answer 2

$\text{\purple{\huge{Solution}}}$

Angle Inscribe Theorem:

The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle.

∴ center angle, ∠ AOB = 2 × ∠ ACB

Substituting we get

$\sf{\angle{AOB = 2 \times 50{\degree} = 100{\degree}}}$

Now, In ΔAOB

AO=BO---------------(Radius of same circle)

∴ ΔAOB is an Isosceles Triangle

∴ ∠OAB = ∠OBA -----------------(Isosceles Triangle Property)

Now, In ΔAOB

$\sf{\angle{AOB + {\angle{OAB + {\angle{OBA = 180{\degree}}}}}}}$

Substituting we get

$\sf{2m{\angle{OAB = 180 - 100 = 80{\degree}}}}$

$\sf{m{\angle{OAB = \frac{80}{2} = 40{\degree}}}}$

$\mathcal{\blue{\boxed{Hence,\:the\:answer\:is\:40{\degree}}}}$