Question

In the figure O is the centre of the circle . Central angle of arc AXB is 60 degree and arc CYD is 80 degree .find the all angle of triangle APD ?​

Answer 1

Answer:

There isn't any figure........

Step-by-step explanation:

I KNOW I HAVEN'T HELPED YOU BUT THEN ALSO MARK ME AS BRAINIEST

Answer 2

Answer:

Given:

In given figure, AOC is a diameter of the circle

arc AXB = \frac{1}{2}\times

2

1

× arc BYC

We know that, central angle is always equal to its opposite arc

Therefore ∠AOB = \frac{1}{2}\times

2

1

× ∠BOC (1)

Also, ∠ AOB + ∠BOC = 180° [linear pair axiom]

\frac{1}{2}\times \angle BOC + \angle BOC = 180\°

2

1

×∠BOC+∠BOC=180\° [from equation 1]

\frac{\angle BOC +2 \angle BOC}{2}=180\°

2

∠BOC+2∠BOC

=180\°

\frac{3 \angle BOC}{2} =180\°

2

3∠BOC

=180\°

∠BOC = \frac{180\times2}{3}

3

180×2

∠BOC = \frac{360}{3}

3

360

Hence ∠BOC = 120°.

I have given u example I hope now u can do it