Question

1. In Fig. 15.120, O is the centre of the circle. If ZAPB = 50°, find ZAOB and ZOAB.​

Answer 1

∠ OAB = ∠ OBA = 40°

Step-by-step explanation:

Given:

O is the centre of the circle.

∠ APB = 50°

To find:

∠ AOB and ∠ OAB.​

Solution:

O is the centre of the circle. Let OA and OB be the radius of the circle.

∴ OA = OB

∴ ∠ OAB = ∠ OBA            [Angles opposite to equal sides]      [1]

Also,

∠ AOB = 2 ∠ APB             [Degree measure theorem]

$\angle OAB = 2 \times 50$

∠ AOB = 100°                          [2]

Now let us assume that, ∠ OAB = x

So in Δ OAB,

∠ AOB + ∠ OAB + ∠ OBA  = 180°    [Angle sum property]

⇒ x +x + 100° = 180°                      [from 2]

⇒ 2x  = 180° - 100°

⇒ 2x = 80°

⇒ $x=\frac{80}{2}$

⇒ x = 40°

So therefore  from Eq (1), we get

∠ OAB = ∠ OBA = 40°

 

 

Answer 2

$\angle AOB$ = 100° and $\angle OAB$ = 40°.

Step-by-step explanation:

We are given that O is the center of the circle and $\angle$APB = 50°.

Now, as shown in the figure the $\angle$APB = 50°, so due to degree measure theorem it is stated that;

$\angle$AOB = 2 $\times$ $\angle$APB     {because of degree measure theorem}

$\angle$AOB = 2 $\times$ 50°

So, $\angle$AOB = 100°

Now, as we can see in the figure that OA = OB {because they are the radius of the circle so they must be same}.

This means that $\angle$OAB = $\angle$OBA  {because equal sides have equal opposite angles}

Let the $\angle$OAB = x , so $\angle$OBA will also be x.

Now, considering $\triangle$AOB, we know that the sum of angles of a triangle is 180°, i.e;

       $\angle$AOB + $\angle$OAB + $\angle$OBA = 180°

         100° + x + x = 180°

             2x = 180° - 100°

              2x = 80°

                x = $\frac{80}{2}$ = 40°

This means that $\angle$OAB = $\angle$OBA = 40°.