Question

AB is a tangent to a circle with centre O at point A. If ∆OAB is an isosceles

triangle, then find OBA.

​

Answer 1

ANSWER

In △OAB,

OA=AB ....given

∴△OAB is isosceles.

∴∠OBA=∠AOB=x ....(∵OA=AB)

∠OAB+∠ABO+∠AOB=180

o

...Angle sum property o triangle

∴90

∘

+x+x=180

∘

⇒x=45

∘

So, option B is correct.

Answer 2

Answer:∠OBA = 45°

Step-by-step explanation:

Concept: A line that touches the circle at a single point is known as a tangent to a circle. The point where tangent meets the circle is called point of tangency. The tangent is perpendicular to the radius of the circle, with which it intersects. Tangent can be considered for any curved shapes.

Step 1: ΔOAB is an isosceles triangle, where O is the center of the circle, and B is the tangent to the circle with center O at point A.

Step 2: We know that the tangent to a circle always forms a right angle .

So, ∠OAB= 90°.

Step 3: Since ΔOAB is an isosceles triangle and one angle of the triangle is 90°, hence we can conclude that the other two angles are equal (property of an isosceles triangle). So ∠AOB = ∠ABO= $x$ (say). So automatically, from the property of an isosceles triangle , the two sides , AO and AB are equal in length.

Step 4: Now by the angle sum property of a triangle, we know that the sum of all the interior angles of a triangle equals 180°.

So, in the above traingle, 90° + $2x$ = 180°

                                        ⇒$2x$ = 90°

                                         ⇒ $x$ = 45°

Final Ans: ∠OBA = 45°