Question

In the given figure, XY and XY are two parallel tangents to a circle with centre O and another tangent AB with point of contact C, is intersecting XY at A and XY at B. Prove that  AOB = 90.

Answer 1

Correct question:-

In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.

$\huge\star{\green{\underline{\mathfrak{Explanation:}}}}$

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Now the triangles △OPA and △OCA are similar using SSS congruency as:

• OP = OC -They are the radii of the same circle

• AO = AO - It is the common side

• AP = AC - These are the tangents from point A

So, △OPA ≅ △OCA

Similarly,

△ OQB ≅ △ OCB

So,

∠POA = ∠COA …...... (Equation i)

And, ∠QOB = ∠COB …....... (Equation ii)

Since the line POQ is a straight line, it can be considered as a diameter of the circle.

So, ∠ POA + ∠ COA + ∠ COB + ∠ QOB = 180°

Now, from equations (i) and equation (ii) we get that,

2∠ COA + 2∠ COB = 180°

⇒ ∠ COA + ∠ COB = 90°

∴ ∠ AOB = 90°

$\huge\star{\red{\underline{\underline{\mathfrak{Hence,proved!}}}}}$

Answer 2

$\huge\boxed{\fcolorbox{lime}{black}{⭐ANSWER~~}}$

From the figure given in the textbook, join OC. Now, the diagram will be as-

Ncert solutions class 10 chapter 10-11

Now the triangles △OPA and △OCA are similar using SSS congruency as:

(i) OP = OC They are the radii of the same circle

(ii) AO = AO It is the common side

(iii) AP = AC These are the tangents from point A

So, △OPA ≅ △OCA

Similarly,

△OQB ≅ △OCB

So,

∠POA = ∠COA … (Equation i)

And, ∠QOB = ∠COB … (Equation ii)

Since the line POQ is a straight line, it can be considered as a diameter of the circle.

So, ∠POA +∠COA +∠COB +∠QOB = 180°

Now, from equations (i) and equation (ii) we get,

2∠COA+2∠COB = 180°

∠COA+∠COB = 90°

∴∠AOB = 90°