Question

in the given figure O is the centre of a circle in which angle OAB=20° and angle OCB=50° then find angle AOC

Answer 1

Answer:

∠AOC=60°

Step-by-step explanation:

In the Δ OAB

OA=OB=r

So ∠OBA=∠OAB=20°

By angle sum property in ΔOAB

∠OBA+∠OAB+∠AOB=180

20+20+∠AOB=180

∠AOB=180-20-20=140................(1)

Now in the ∠OBC, OB=OC=r

∠OBC=∠OCB=50°

By angle sum property in ΔOAB

∠OBC+∠OCB+∠BOC=180

50+50+∠BOC=180

THUS ∠BOC=180-100=80°

but ∠AOB=140°

So ∠AOC +∠BOC=140

or ∠AOC +80=140

∠AOC=140-80=60

Thus ∠AOC=60°

Answer 2

The value of angle AOC is (d) 60°.

Given:

∠$OAB=20^{o}$

∠$OCB=50^{o}$

To find:

∠$AOC$

Solution:

Step 1

In the given figure, we can see that since $OA$ and $OB$ are the radius of the same circle, hence,

$OA=OB$

Also, we know the angles opposite to the same side of the triangle are equal therefore, we also get

∠$OBA=$ ∠$OAB$

We know the value of ∠$OAB=20^{o}$. Hence, we get

∠$OBA=20^{o}$

Now,

In Δ$OAB$, applying angle sum property of the triangle, we get

∠$OAB$ $+$ ∠$OBA$ $+$ ∠$AOB=180^{o}$

$20^{o}+ 20^{o}$ $+$ ∠$AOB=180^{o}$

∠$AOB=140^{o}$

Step 2

Now,

In the diagram we can also see that $OB$ and $OC$ are the radius of the same circle, hence,

$OB=OC$

Angles opposite to the same sides of the triangle are equal. Hence,

∠$OCB=$ ∠$OBC$

∠$OBC=50^{o}$

Now,

In Δ$BOC$, applying the angle sum property of triangle, we get

∠$OCB$ $+$ ∠$OBC$ $+$ ∠$BOC=180^{o}$

We know, ∠$OCB=50^{o}$

$50^{o}+ 50^{o}+$ ∠$BOC=180^{o}$

∠$BOC=80^{o}$

Step 3

Now,

From the figure, we can see that

∠$AOB=$ ∠$AOC+$ ∠$BOC$

We know, ∠$AOB=140^{o}$ and ∠$BOC=80^{o}$

Substituting the known vales, we get

$140^{o}=$ ∠$AOC+80^{o}$

∠$AOC=60^{o}$

Final answer:

Hence, the value of ∠AOC is (d) 60°.

Although your question is incomplete, you might be referring to the question below.

In the given figure, O is he center of a circle in which ∠OAB = 20° and ∠OCB = 50°, then, ∠AOC = ?

a) 50°

b) 70°

c) 20°

d) 60°