In the figure, given below, O is the centre of the circle. If ∠AOB = 140degree and ∠OAC = 50degree; Find (i) ∠OAB, (ii) ∠CBA. - Noon Academy

In the figure, given below, O is the centre of the circle. If ∠AOB = 140degree and ∠OAC = 50degree; Find (i) ∠OAB, (ii) ∠CBA.

Circles | Class 10 | Exercise 17A | ICSE | Maths | Selina

In the figure, given below, O is the centre of the circle. If ∠AOB = 140degree and ∠OAC = 50degree; Find (i) ∠OAB, (ii) ∠CBA.

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

Solution:

According to the given question,

(i) In

$\vartriangle AOB,$

we have

$OA\text{ }=\text{ }OB\text{ }\left( radii \right)$

Thus,

$\angle OBA\text{ }=\angle OAB$

By angle sum property of a triangle

$\angle OBA\text{ }+\angle OAB\text{ }+\angle AOB\text{ }=\text{ }{{180}^{o}}$

$2\angle OBA\text{ }+\text{ }{{140}^{o}}~=\text{ }{{180}^{o}}$

$2\angle OBA\text{ }=\text{ }{{40}^{o}}$

$\angle OBA\text{ }=\text{ }{{20}^{o}}$

(ii) We know that,

$\angle OBC\text{ }=\text{ }{{60}^{o}}$

And,

$\angle OBC\text{ }=\angle CBA\text{ }+\angle OBA$

${{60}^{o}}~=\angle CBA\text{ }+\text{ }{{20}^{o}}$

Therefore,

$\angle CBA\text{ }=\text{ }{{40}^{o}}$

i

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