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Q4:
In each of the following, give also the justification of the construction:
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of $60^{\circ}$.
Solution :
Given: A circle with center $O$ and radius $r = 5\text{ cm}$. The angle between the two tangents is $60^{\circ}$.
To Find: Construct the pair of tangents and provide the geometric justification for the construction.
Step 1: Conceptual Analysis
Let the circle have center $O$. Let the two tangents meet at point $P$. Let the points of contact be $A$ and $B$.
In the quadrilateral $OAPB$:
- $\angle OAP = 90^{\circ}$ (Radius is perpendicular to the tangent at the point of contact).
- $\angle OBP = 90^{\circ}$ (Radius is perpendicular to the tangent at the point of contact).
- $\angle APB = 60^{\circ}$ (Given).
- The sum of angles in a quadrilateral is $360^{\circ}$.
Therefore, $\angle AOB = 360^{\circ} - (90^{\circ} + 90^{\circ} + 60^{\circ}) = 360^{\circ} - 240^{\circ} = 120^{\circ}$.
Step 2: Construction Procedure
1. Draw a circle of radius $5\text{ cm}$ with center $O$.
2. Draw any radius $OA$.
3. Construct an angle of $120^{\circ}$ at the center $O$ such that $\angle AOB = 120^{\circ}$.
4. At point $A$, construct a perpendicular to $OA$.
5. At point $B$, construct a perpendicular to $OB$.
6. Let these two perpendiculars intersect at point $P$. $PA$ and $PB$ are the required tangents.
Step 3: Justification
In quadrilateral $OAPB$:
- $\angle OAP = 90^{\circ}$ [By construction, as the tangent is perpendicular to the radius].
- $\angle OBP = 90^{\circ}$ [By construction, as the tangent is perpendicular to the radius].
- $\angle AOB = 120^{\circ}$ [By construction].
- The sum of interior angles of a quadrilateral is $360^{\circ}$.
- $\angle APB = 360^{\circ} - (\angle OAP + \angle OBP + \angle AOB)$
- $\angle APB = 360^{\circ} - (90^{\circ} + 90^{\circ} + 120^{\circ})$
- $\angle APB = 360^{\circ} - 300^{\circ} = 60^{\circ}$.
Thus, the tangents are inclined to each other at $60^{\circ}$.
Final Answer: The construction is justified as the angle between the tangents is calculated to be $60^{\circ}$ based on the properties of the quadrilateral $OAPB$ formed by the radii and the tangents.
More Questions from Class 10 Mathematics Constructions EXERCISE 11.2
- Q1: In each of the following, give also the justification of the construction: Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
- Q2: In each of the following, give also the justification of the construction: Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
- Q3: In each of the following, give also the justification of the construction: Draw a circle of radius 3 cm. Take two points $P$ and $Q$ on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points $P$ and $Q$.
- Q5: In each of the following, give also the justification of the construction: Draw a line segment $AB$ of length 8 cm. Taking $A$ as centre, draw a circle of radius 4 cm and taking $B$ as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
- Q6: In each of the following, give also the justification of the construction: Let ABC be a right triangle in which $AB = 6$ cm, $BC = 8$ cm and $\angle B = 90^{\circ}$. $BD$ is the perpendicular from $B$ on $AC$. The circle through $B$, $C$, $D$ is drawn. Construct the tangents from $A$ to this circle.
- Q7: In each of the following, give also the justification of the construction: Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
CBSE Solutions for Class 10 Mathematics Constructions
Chapters in CBSE - Class 10 Mathematics
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