In the given figure, ABC is a triangle in which ∠BAC = 30o. Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose center is O.
In the given figure, ABC is a triangle in which ∠BAC = 30o. Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose center is O.

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(C) - 2

Solution:

Firstly, join

    \[OB\text{ }and\text{ }OC\]

Proof:

    \[\angle BOC\text{ }=\text{ }2\angle BAC\text{ }=\text{ }2\text{ }x\text{ }{{30}^{o}}~=\text{ }{{60}^{o}}\]

Now, in

    \[\vartriangle OBC\]

    \[OB\text{ }=\text{ }OC\]

[Radii of same circle]

So,

    \[\angle OBC\text{ }=\angle OCB\]

[Angles opposite to equal sides]

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(C) - 3

And in

    \[\vartriangle OBC\]

by angle sum property we have

    \[\angle OBC\text{ }+\angle OCB\text{ }+\angle BOC\text{ }=\text{ }{{180}^{o}}\]

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

Or,

    \[2\angle OBC\text{ }=\text{ }{{180}^{o}}-\text{ }{{60}^{o}}~=\text{ }{{120}^{o}}\]

    \[\angle OBC\text{ }=\text{ }{{120}^{o}}/\text{ }2\text{ }=\text{ }{{60}^{o}}\]

So,

    \[\angle OBC\text{ }=\angle OCB\text{ }=\angle BOC\text{ }=\text{ }{{60}^{o}}\]

Thus,

    \[\vartriangle OBC\]

is an equilateral triangle.

So,

    \[BC\text{ }=\text{ }OB\text{ }=\text{ }OC\]

But,

    \[OB\text{ }and\text{ }OC\]

are the radii of the circum-circle.

Therefore,

    \[BC\]

is also the radius of the circum-circle.