Ex.10.2 Q9 Circles Solution - NCERT Maths Class 10

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Question

In Figure, \(XY\) and \(X'Y' \) are two parallel tangents to a circle with center \(O\) and another tangent \(AB\) with point of contact \(C\) intersecting \(XY\) at \(A\) and \(X'Y' \) at \(B.\)

Prove that \(\angle {AOB} = {90^ \circ }\).

 Video Solution
Circles
Ex 10.2 | Question 9

Text Solution

What is known?

  •  '\(O\)' is the centre of the circle.
  • \(XY\) and \(X'Y'\)' are the two parallel tangents to the circle.
  • \(AB\) is another tangent with point of contact \(C\), intersecting \(XY\) at \(A\) and \(X'Y'\) at \(B.\)

To prove:

\(\angle {AOB} = {90^ \circ }\)

Reasoning :

Join point \(O\) to \(C.\)

In \(\Delta OPA\) and \(\Delta OCA\)

\(OP = OC\) (Radii of the circle are equal)

\(AP = AC \) (The lengths of tangents drawn from an external point \(A\) to a circle are equal.)

\(AO = AO\)  (Common)

By \(SSS\) congruency, \(\Delta {OPA} \cong \Delta {OCA}\)

SSS Congruence Rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Steps:

Therefore, \(\angle {OPA} = \angle {OCA}\)

Similarly, \(\Delta {OCB} \cong \Delta {OBQ}\)

Therefore, \(\angle {COB} = \angle {BOQ}\)

\(PQ\) is a diameter, hence a straight line and \(\angle {POQ} = {180^ \circ}\)

But \(\begin{align} \angle {P O Q} =\begin{bmatrix} \angle {P O A} + \angle {A O C}+ \\ \angle {C O B} + \angle {B O Q}\end{bmatrix}\end{align}\)

Therefore,
\[\begin{align}  & \left[ \begin{array}  \angle POA+\angle AOC+ \\  \angle COB+\angle BOQ \\ \end{array} \right]={{180}^{{}^\circ }} \\ &  2\angle AOC+ 2\angle COB ={{180}^{{}^\circ }}\end{align}\]

Therefore,
\[\begin{align}& \left[ \begin{array}& \angle POA=\angle AOC \;\text{and}  \\\angle COB=\angle BOQ \\ \end{array} \right] \end{align}\]

Therefore,
\[\angle AOC+\angle COB={{90}^{{}^\circ }}\]

From the figure, \(\angle {A O C} + \angle {C O B} = \angle {A O B}\)

\(\therefore \quad \angle {A O B} = 90 ^ { \circ }\)

Hence Proved

  
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