Question

In the figure above (not to scale), $AB$ is the diameter of the circle with centre $O$. If $\angle ACO={30}^o$, the $\angle BOC$.

• A. ${60}^o$
• B. ${50}^o$
• C. ${55}^o$
• D. ${45}^o$

Answer 1

Answer: (A)

${60}^o$

given that,

$\angle ACO={30}^0$

Then, we know that

$\angle ACB={90}^0$ (angle formed in semi circle is a right angle)

Now,

$\angle ACB=\angle ACO+\angle BCO$

${90}^0={30}^0+\angle BCO$

$\angle BCO={60}^0$

In $\Delta AOC$

We know that.

$AO=CO$ (radius of circle)

Then,

$\angle OAC=\angle ACO$

$\angle OAC={30}^0$

In $\Delta BOC$

We know that.

$BO=CO$ (radius of circle)

Then,

$\angle OBC=\angle BCO$

$\angle OBC={60}^0$

Again,In $\Delta BOC$

We know that,

$\angle BOC+\angle OBC+\angle BCO={180}^0$

$\angle BOC+{60}^0+{60}^0={180}^0$

$\angle BOC={180}^0-{120}^0$

$\angle BOC={60}^0$

Hence, this is the answer.