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Q2:
Choose the correct option and give justification. In Fig. 10.11, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle POQ = 110^{\circ}$, then $\angle PTQ$ is equal to
Choose the correct option and give justification. In Fig. 10.11, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle POQ = 110^{\circ}$, then $\angle PTQ$ is equal to
Solution :
Given:
A circle with centre $O$. $TP$ and $TQ$ are two tangents drawn from an external point $T$ to the circle. The angle between the radii at the points of contact is $\angle POQ = 110^{\circ}$.
To Find:
The measure of $\angle PTQ$.
Visual Representation:
Step 1: Identifying Geometric Properties
According to the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."
Therefore, $OP \perp TP$ and $OQ \perp TQ$.
This implies that $\angle OPT = 90^{\circ}$ and $\angle OQT = 90^{\circ}$.
Step 2: Analyzing the Quadrilateral
Consider the quadrilateral $OPTQ$. The sum of the interior angles of a quadrilateral is always $360^{\circ}$.
The sum of the angles is given by:
$\angle POQ + \angle OPT + \angle PTQ + \angle OQT = 360^{\circ}$
Step 3: Substituting Known Values
Substitute the known values into the equation:
$110^{\circ} + 90^{\circ} + \angle PTQ + 90^{\circ} = 360^{\circ}$
Step 4: Solving for $\angle PTQ$
Combine the constant terms:
$110^{\circ} + 180^{\circ} + \angle PTQ = 360^{\circ}$
$290^{\circ} + \angle PTQ = 360^{\circ}$
Subtract $290^{\circ}$ from both sides:
$\angle PTQ = 360^{\circ} - 290^{\circ}$
$\angle PTQ = 70^{\circ}$
Final Answer: The measure of $\angle PTQ$ is $70^{\circ}$.
More Questions from Class 10 Mathematics Circles EXERCISE 10.2
- Q1: Choose the correct option and give justification. From a point $Q$, the length of the tangent to a circle is $24$ cm and the distance of $Q$ from the centre is $25$ cm. The radius of the circle is
- Q10: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
- Q11: Prove that the parallelogram circumscribing a circle is a rhombus.
- Q12: A triangle $ABC$ is drawn to circumscribe a circle of radius $4$ cm such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8$ cm and $6$ cm respectively (see Fig. 10.14). Find the sides $AB$ and $AC$.
- Q13: Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
- Q3: Choose the correct option and give justification. If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of $80^{\circ}$, then $\angle POA$ is equal to
- Q4: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
- Q5: Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
- Q6: The length of a tangent from a point $A$ at distance $5$ cm from the centre of the circle is $4$ cm. Find the radius of the circle.
- Q7: Two concentric circles are of radii $5$ cm and $3$ cm. Find the length of the chord of the larger circle which touches the smaller circle.
- Q8: A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig. 10.12). Prove that $AB + CD = AD + BC$.
- Q9: 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 $\angle AOB = 90^{\circ}$.
CBSE Solutions for Class 10 Mathematics Circles
Chapters in CBSE - Class 10 Mathematics
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