In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30degree, find: (i) angle BCO (ii) angle AOB
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30degree, find: (i) angle BCO (ii) angle AOB

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(C) - 21

Solution:

In the given fig,

    \[O\]

is the centre of the circle and,

    \[CA\text{ }and\text{ }CB\]

are the tangents to the circle from

    \[C\]

Also, 

    \[\angle ACO\text{ }=\text{ }{{30}^{o}}\]

    \[P\]

is any point on the circle.

    \[P\text{ }and\text{ }B\]

are joined.

To find:

    \[\left( i \right)\angle BCO~\]

    \[\left( ii \right)~\angle AOB\]

 

Proof:

(i) In

    \[\vartriangle OAC\text{ }and\text{ }\vartriangle OBC\]

we have

    \[OC\text{ }=\text{ }OC\]

[Common]

    \[OA\text{ }=\text{ }OB\]

[Radii of the same circle]

    \[CA\text{ }=\text{ }CB\]

[Tangents to the circle]

Hence,

    \[\vartriangle OAC\cong \vartriangle OBC\text{ }by\text{ }SSS\]

congruence criterion

Thus,

    \[\angle ACO\text{ }=\angle BCO\text{ }=\text{ }{{30}^{o}}\]

 

(ii) As

    \[\angle ACB\text{ }=\text{ }{{30}^{o}}~+\text{ }{{30}^{o}}~=\text{ }{{60}^{o}}\]

And,

    \[\angle AOB\text{ }+\angle ACB\text{ }=\text{ }{{180}^{o}}\]

    \[\angle AOB\text{ }+\text{ }{{60}^{o}}~=\text{ }{{180}^{o}}\]

    \[\angle AOB\text{ }=\text{ }{{180}^{o}}-\text{ }{{60}^{o}}\]

So,

    \[\angle AOB\text{ }=\text{ }{{120}^{o}}\]