Question

In the figure O is the center and AB is a diameter. If <A = 30⁰.

(a) Compute < ACO​

Answer 1

Answer:

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Answer 2

given that,

∠ACO=300

Then, we know that

∠ACB=900     (angle formed in semi circle is a right angle)

 

Now,

∠ACB=∠ACO+∠BCO

900=300+∠BCO

∠BCO=600

 

In ΔAOC

We know that.

AO=CO       (radius of circle)

Then,

∠OAC=∠ACO

∠OAC=300

 

In ΔBOC

We know that.

BO=CO       (radius of circle)

Then,

∠OBC=∠BCO

∠OBC=600

 

Again,In ΔBOC

We know that,

∠BOC+∠OBC+∠BCO=