Question

AB is a radius of a circle with centre O . C is a point on the circle. If angle OBC = 60° then find the value of angle OCA
please give right answer​

Answer 1

Answer:

i think the answer is 0 degreeas because O andA lies in the same point

Answer 2

$\Large\bf\blue{Given : } \:$

AB is a diameter of a circle with centre O . C is a point on the circle.

$\Large\bf\blue{ To \: \: find: } \:$

$\large\mathrm{Value \: of \: \angle {OCA}} \:$

$\Large\bf\blue{ \: Solution : } \:$

In the figure,

$\mathrm{AO = OB ( radius \: of \: the \: circle) } \:$

$\mathrm{AB \: is \: the \: Diameter \: of \: the \: circle} \:$

$\mathrm{ In \: \: \: \triangle{BOC,}} \:$

$\mathrm{ \angle{BOC} = 90^{\circ}} \:$

we know that,

Sum of all angles of triangle = 180°

ie.,

$\mathrm{\angle{BOC} + \angle{ OBC}+ \angle{OCB}= 180^\circ} \:$

$\implies\mathrm{90^\circ + 60^\circ + \angle{OCB} = 180^\circ} \\ \\ \implies \mathrm{150^\circ + \angle{OCB} = 180^\circ} \\ \\ \implies \mathrm{\angle{OCB} = 180^\circ - 150^\circ} \\ \\ \implies \fbox{\mathrm{\angle{OCB} = 30^\circ}} \:$

Now,

In the given figure :

$\mathrm{\angle{ACB} = 60 ^\circ} \:$

$\rightarrow\mathrm{ \angle{ACO} + \angle {BCO \: = 60 \degree}} \:$

$\rightarrow\mathrm{\angle{ACO} + 30^\circ = 60^\circ }\:$

$\rightarrow\mathrm{\angle{ACO} = 60^\circ - </p><p>30^\circ} \:$

$\rightarrow\fbox\mathrm{\angle{ACO} = 30^\circ} \:$

$\huge\mathfrak\purple{Solved.} \: \: \: \:$