Mix Examples - Constructions Questions in English

(A) To divide a line segment $AB$ in the ratio $p:q$,we draw a ray $AX$ making an acute angle with $AB$.
We then mark $p+q$ points on the ray $AX$ at equal distances,say $A_1, A_2, \dots, A_{p+q}$,such that $AA_1 = A_1A_2 = \dots = A_{p+q-1}A_{p+q}$.
We join the last point $A_{p+q}$ to $B$.
Then,we draw a line parallel to $A_{p+q}B$ through the point $A_p$ to intersect $AB$ at a point $P$.
This point $P$ divides the line segment $AB$ in the ratio $p:q$.
Therefore,the minimum number of points to be marked on the ray $AX$ is $p+q$.

(B) Let the angle between the two tangents be $\theta = 35^{\circ}.$
In the quadrilateral formed by the center of the circle,the two points of contact,and the intersection point of the tangents,the angles at the points of contact are $90^{\circ}$ each (since the radius is perpendicular to the tangent).
Thus,the sum of the angles in the quadrilateral is $360^{\circ}.$
Let the angle between the radii be $\alpha.$
Then,$\alpha + 90^{\circ} + \theta + 90^{\circ} = 360^{\circ}.$
$\alpha + \theta = 180^{\circ}.$
$\alpha = 180^{\circ} - 35^{\circ} = 145^{\circ}.$

(C) We know that,to divide a line segment $AB$ in the ratio $m:n,$ first draw a ray $AX$ which makes an acute angle $\angle BAX$,then mark $m+n$ points at equal distances.
Here,$m=5$ and $n=7$.
So,the minimum number of these points $= m+n = 5+7 = 12$.

(D) To divide a line segment in the ratio $m:n,$ we first draw a ray making an acute angle with the segment.
Then,we mark $m+n$ points at equal distances on the ray.
Here,the ratio is $4:7,$ so the total number of points required is $4+7 = 11$.
Therefore,we mark points $A_1, A_2, \dots, A_{11}$ on the ray $AX$.
The last point $A_{11}$ is joined to point $B$ to complete the construction.

(A) To divide a line segment $AB$ in the ratio $m: n$ (here $5: 6$),we follow these steps:
$1$. Draw a ray $AX$ making an acute angle $\angle BAX$ with $AB$.
$2$. Draw a ray $BY$ parallel to $AX$ by making $\angle ABY = \angle BAX$.
$3$. Locate $m$ points $(A_{1}, A_{2}, \ldots, A_{5})$ on $AX$ and $n$ points $(B_{1}, B_{2}, \ldots, B_{6})$ on $BY$ such that $AA_{1} = A_{1}A_{2} = \ldots = A_{4}A_{5}$ and $BB_{1} = B_{1}B_{2} = \ldots = B_{5}B_{6}$.
$4$. Join the point $A_{m}$ (which is $A_{5}$) to the point $B_{n}$ (which is $B_{6}$). The line segment $A_{5}B_{6}$ intersects $AB$ at point $C$,dividing $AB$ in the ratio $5: 6$.

(B) To construct a triangle similar to a given $\triangle ABC$ with a scale factor of $\frac{3}{7}$,we follow these steps:
$1$. Draw a ray $BX$ making an acute angle with $BC$.
$2$. Locate $7$ points $(B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}, B_{7})$ on $BX$ at equal distances because the denominator of the fraction $\frac{3}{7}$ is $7$.
$3$. The next step is to join the last point,$B_{7}$,to $C$ to establish the base line for the construction.

(C) To construct a triangle similar to a given triangle with a scale factor of $\frac{m}{n}$,where $m$ and $n$ are positive integers,the number of points to be marked on the ray $BX$ is equal to the greater of $m$ and $n$.
In this problem,the scale factor is $\frac{8}{5}$.
Comparing the numerator $m = 8$ and the denominator $n = 5$,the greater value is $8$.
Therefore,the minimum number of points to be located at equal distances on ray $BX$ is $8$.

(D) Let the center of the circle be $O$ and the intersection point of the two tangents be $P$. Let the points of contact of the tangents be $A$ and $B$.
In the quadrilateral $OAPB$,the angles at the points of contact $A$ and $B$ are $90^{\circ}$ each (since the tangent is perpendicular to the radius at the point of contact).
The sum of the angles in a quadrilateral is $360^{\circ}$.
Therefore,$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^{\circ}$.
Substituting the known values: $\angle AOB + 90^{\circ} + 60^{\circ} + 90^{\circ} = 360^{\circ}$.
$\angle AOB + 240^{\circ} = 360^{\circ}$.
$\angle AOB = 360^{\circ} - 240^{\circ} = 120^{\circ}$.
Thus,the angle between the two radii should be $120^{\circ}$.

(B) False.
The ratio $2 \sqrt{3} : 2 \sqrt{3}$ simplifies to $1 : 1$.
However,the standard geometrical construction method for dividing a line segment in a given ratio $m : n$ requires $m$ and $n$ to be positive integers.
Since $2 \sqrt{3}$ is an irrational number and not a positive integer,it is not possible to divide a line segment in this ratio using the standard ruler and compass construction method.

(A) True.
Given ratio $= \sqrt{3} : \frac{1}{\sqrt{3}}$.
To simplify this ratio,multiply both terms by $\sqrt{3}$:
$\sqrt{3} \times \sqrt{3} : \frac{1}{\sqrt{3}} \times \sqrt{3} = 3 : 1$.
Since $3 : 1$ is a ratio of two positive integers,it is always possible to divide a line segment in this ratio using a ruler and compass by constructing $3 + 1 = 4$ equal parts on an auxiliary ray.
Therefore,the statement is True.

(B) False.
To construct a triangle similar to $\triangle ABC$ with a scale factor of $\frac{7}{3}$,we must divide the ray $BX$ into $7$ equal parts (since $7 > 3$).
Steps of construction:
$1.$ Draw a ray $BX$ making an acute angle with $BC$.
$2.$ Locate $7$ points $B_{1}, B_{2}, \dots, B_{7}$ on $BX$ such that $BB_{1} = B_{1}B_{2} = \dots = B_{6}B_{7}$.
$3.$ Join $B_{3}$ to $C$ (since the denominator is $3$).
$4.$ Draw a line through $B_{7}$ parallel to $B_{3}C$ intersecting the extended line segment $BC$ at $C'$.
$5.$ Draw a line through $C'$ parallel to $AC$ intersecting the extended line segment $BA$ at $A'$.
The given statement claims that $B_{6}C'$ is drawn parallel to $B_{3}C$,which is incorrect because the scale factor is $\frac{7}{3}$,requiring the parallel line to be drawn from $B_{7}$,not $B_{6}$.

(B) False.
The radius of the circle is $r = 3.5 \, cm$ and the distance of point $P$ from the centre is $d = 3 \, cm$.
Since $d < r$ $(3 \, cm < 3.5 \, cm)$,the point $P$ lies inside the circle.
$A$ tangent to a circle is a line that touches the circle at exactly one point. If a point lies inside the circle,no tangent can be drawn from that point to the circle.

(A) True.
In a circle,the angle between the two tangents drawn from an external point and the angle subtended by the line segments joining the points of contact to the center are supplementary (i.e.,they add up to $180^{\circ}$).
Let the angle between the tangents be $\theta$ and the angle at the center be $\phi$.
We know that $\theta + \phi = 180^{\circ}$.
Since the angle between the tangents is given as $170^{\circ}$,the angle at the center would be $180^{\circ} - 170^{\circ} = 10^{\circ}$.
Since $10^{\circ} > 0^{\circ}$,it is geometrically possible to construct such a pair of tangents.

(A) $1$. First,draw an equilateral triangle $ABC$ where $AB = BC = CA = 4\, cm$.
$2$. Draw a ray $AX$ making an acute angle with $AB$ on the side opposite to vertex $C$.
$3$. Locate $5$ points $A_1, A_2, A_3, A_4, A_5$ on $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5$.
$4$. Join $A_5$ to $B$. Draw a line through $A_3$ parallel to $A_5B$ intersecting $AB$ at $B'$.
$5$. Draw a line through $B'$ parallel to $BC$ intersecting $AC$ at $C'$.
$6$. The triangle $AB'C'$ is the required triangle similar to $\triangle ABC$ with a scale factor of $\frac{3}{5}$.
$7$. Since the ratio of corresponding sides is equal,the angles remain the same ($60^{\circ}$ each). Therefore,the new triangle is also equilateral.

(N/A) $1.$ Draw a line segment $AB = 7 \, cm$.
$2.$ Draw a ray $AX$ making an acute angle $\angle BAX$.
$3.$ Along $AX,$ mark $3 + 5 = 8$ points $A_1, A_2, A_3, A_4, A_5, A_6, A_7, A_8$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7 = A_7A_8$.
$4.$ Join $A_8B$.
$5.$ From $A_3,$ draw $A_3P \parallel A_8B$ meeting $AB$ at $P$ (by making an angle equal to $\angle BA_8A$ at $A_3$).
Then,$P$ is the point on $AB$ which divides it in the ratio $3: 5.$
Thus,$AP: PB = 3: 5.$
Justification:
Let $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = \dots = A_7A_8 = x.$
In $\triangle ABA_8,$ we have $A_3P \parallel A_8B.$
By Basic Proportionality Theorem,$\frac{AP}{PB} = \frac{AA_3}{A_3A_8} = \frac{3x}{5x} = \frac{3}{5}.$
Hence,$AP: PB = 3: 5.$

(A) Steps of construction:
$1.$ Draw a line segment $BC = 12 \, cm$.
$2.$ From point $B$,draw a line $AB = 5 \, cm$ perpendicular to $BC$ such that $\angle B = 90^{\circ}$.
$3.$ Join $AC$. Thus,$\triangle ABC$ is the given right triangle.
$4.$ From $B$,draw an acute angle $\angle CBY$ downwards.
$5.$ On ray $BY$,mark three points $B_1, B_2$ and $B_3$ such that $BB_1 = B_1B_2 = B_2B_3$.
$6.$ Join $B_3C$.
$7.$ From point $B_2$,draw $B_2N \parallel B_3C$ to intersect $BC$ at $N$.
$8.$ From point $N$,draw $NM \parallel CA$ to intersect $BA$ at $M$. Thus,$\triangle MBN$ is the required triangle.
$9.$ Since $NM \parallel CA$ and $BC$ is a transversal,the corresponding angles are equal. Therefore,$\angle MNB = \angle ACB$. Since $\angle B = 90^{\circ}$ in both triangles,$\triangle MBN$ is also a right-angled triangle at $B$.

(A) $1.$ Draw a line segment $BC = 6 \, cm$.
$2.$ Taking $B$ and $C$ as centers,draw two arcs of radii $4 \, cm$ and $5 \, cm$ respectively,intersecting each other at $A$.
$3.$ Join $BA$ and $CA$. $\triangle ABC$ is the required triangle.
$4.$ From $B$,draw any ray $BX$ downwards making an acute angle with $BC$.
$5.$ Mark five points $B_1, B_2, B_3, B_4,$ and $B_5$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.
$6.$ Join $B_3C$. From $B_5$,draw $B_5M \parallel B_3C$ intersecting the extended line segment $BC$ at $M$.
$7.$ From point $M$,draw $MN \parallel CA$ intersecting the extended line segment $BA$ at $N$. Then,$\triangle NBM$ is the required triangle whose sides are $\frac{5}{3}$ times the corresponding sides of $\triangle ABC$.

(N/A) Given,a point $M^{\prime}$ is at a distance of $6\, cm$ from the centre of a circle of radius $4\, cm$.
Steps of construction:
$1.$ Draw a circle of radius $4\, cm$. Let the centre of this circle be $O$.
$2.$ Join $OM^{\prime}$ and bisect it. Let $M$ be the mid-point of $OM^{\prime}$.
$3.$ Taking $M$ as the centre and $MO$ as the radius,draw a circle to intersect the circle $(O, 4\, cm)$ at two points,$P$ and $Q$.
$4.$ Join $PM^{\prime}$ and $QM^{\prime}$. $PM^{\prime}$ and $QM^{\prime}$ are the required tangents from $M^{\prime}$ to the circle $C(O, 4\, cm)$.

(N/A) First,draw the rhombus $ABCD$ in which $AB = 4 \, cm$ and $\angle ABC = 60^{\circ}$ as shown in the figure,and join $AC$.
Construct the triangle $AB'C'$ similar to $\triangle ABC$ with a scale factor of $\frac{2}{3}$ as instructed in the Mathematics Textbook for Class $X$.
Finally,draw the line segment $C'D'$ parallel to $CD$ such that $D'$ lies on $AD$.
Since $\triangle AB'C' \sim \triangle ABC$,we have $\frac{AB'}{AB} = \frac{AC'}{AC} = \frac{B'C'}{BC} = \frac{2}{3}$.
Since $C'D' \parallel CD$,$\triangle AD'C' \sim \triangle ADC$. Thus,$\frac{AD'}{AD} = \frac{AC'}{AC} = \frac{C'D'}{CD} = \frac{2}{3}$.
Since $ABCD$ is a rhombus,$AB = BC = CD = AD$.
Therefore,$AB' = B'C' = C'D' = AD' = \frac{2}{3} AB$.
Since all four sides of the quadrilateral $AB'C'D'$ are equal,$AB'C'D'$ is a rhombus.

(D) Given: $AB = 5 \, cm$ and $AC = 7 \, cm$.
Also,$AP = \frac{3}{4} AB$ and $AQ = \frac{1}{4} AC$.
Calculation of lengths:
$AP = \frac{3}{4} \times 5 = 3.75 \, cm$.
$AQ = \frac{1}{4} \times 7 = 1.75 \, cm$.
Construction Steps:
$1$. Draw $AB = 5 \, cm$.
$2$. Draw a ray $AZ$ making an angle of $60^{\circ}$ with $AB$.
$3$. Mark point $C$ on $AZ$ such that $AC = 7 \, cm$.
$4$. To locate $P$ on $AB$ such that $AP = \frac{3}{4} AB$,divide $AB$ in the ratio $3:1$ using the standard division of a line segment method.
$5$. To locate $Q$ on $AC$ such that $AQ = \frac{1}{4} AC$,divide $AC$ in the ratio $1:3$ using the standard division of a line segment method.
$6$. Join $PQ$.
$7$. By measurement,the length $PQ \approx 3.25 \, cm$.

(A) $1.$ Draw a line segment $AB = 3 \, cm$.
$2.$ Draw a ray $BY$ making an angle $\angle ABY = 60^{\circ}$.
$3.$ With $B$ as the center and a radius of $5 \, cm$,draw an arc to cut the ray $BY$ at point $C$.
$4.$ Draw a ray $AZ$ parallel to $BY$ such that $\angle BAZ = 120^{\circ}$ (since $ABCD$ is a parallelogram,$\angle ABC + \angle BCD = 180^{\circ}$).
$5.$ With $A$ as the center and a radius of $5 \, cm$,draw an arc to cut the ray $AZ$ at point $D$.
$6.$ Join $CD$ to complete the parallelogram $ABCD$.
$7.$ Join $BD$,which is a diagonal of the parallelogram $ABCD$.
$8.$ From $B$,draw any ray $BX$ downwards making an acute angle $\angle CBX$.
$9.$ Locate $4$ points $B_1, B_2, B_3, B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$10.$ Join $B_3C$ and draw a line through $B_4$ parallel to $B_3C$ intersecting the extended line segment $BC$ at $C'$.
$11.$ From point $C'$,draw $C'D' \parallel CD$ intersecting the extended line segment $BD$ at $D'$.
$12.$ Draw a line segment $D'A'$ parallel to $DA$,where $A'$ lies on the extended side $BA$.
$13.$ Yes,$A'BCD'$ is a parallelogram because by construction,$A'D' \parallel BC$ and $A'B \parallel D'C'$.

(N/A) Given: Two concentric circles with centre $O$ and radii $3\, cm$ and $5\, cm$. We need to draw a pair of tangents from a point $P$ on the outer circle to the inner circle.
Steps of construction:
$1.$ Draw two concentric circles with centre $O$ and radii $3\, cm$ and $5\, cm$.
$2.$ Take any point $P$ on the outer circle. Join $OP$.
$3.$ Bisect $OP$. Let $M'$ be the midpoint of $OP$.
$4.$ Taking $M'$ as the centre and $OM'$ as the radius,draw a dotted circle which intersects the inner circle at points $M$ and $P'$.
$5.$ Join $PM$ and $PP'$. Thus,$PM$ and $PP'$ are the required tangents.
$6.$ On measuring $PM$ and $PP'$,we find that $PM = PP' = 4\, cm$.
Actual calculation:
In right-angled $\triangle OMP$,$\angle PMO = 90^{\circ}$.
By Pythagoras theorem: $OP^2 = PM^2 + OM^2$
$PM^2 = OP^2 - OM^2$
$PM^2 = (5)^2 - (3)^2 = 25 - 9 = 16$
$PM = \sqrt{16} = 4\, cm$.
Hence,the length of both tangents is $4\, cm$.

(N/A) Let $\triangle PQR$ and $\triangle ABC$ be similar triangles. The scale factor between the corresponding sides is $\frac{PQ}{AB} = \frac{8}{6} = \frac{4}{3}$.
Steps of construction:
$1.$ Draw a line segment $BC = 5 \, cm$.
$2.$ Taking $B$ and $C$ as centers,draw two arcs of radius $6 \, cm$ intersecting each other at $A$.
$3.$ Join $BA$ and $CA$. Thus,$\triangle ABC$ is the required isosceles triangle.
$4.$ From $B$,draw any ray $BX$ making an acute $\angle CBX$.
$5.$ Locate four points $B_1, B_2, B_3,$ and $B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$6.$ Join $B_3C$ and from $B_4$,draw a line $B_4R \parallel B_3C$ intersecting the extended line segment $BC$ at $R$.
$7.$ From point $R$,draw $RP \parallel CA$ meeting the extended line segment $BA$ at $P$.
Then,$\triangle PBR$ is the required triangle.
Justification:
Since $B_4R \parallel B_3C$ (by construction),
$\therefore \frac{BC}{CR} = \frac{3}{1}$.
Now,$\frac{BR}{BC} = \frac{BC + CR}{BC} = 1 + \frac{CR}{BC} = 1 + \frac{1}{3} = \frac{4}{3}$.
Also,$RP \parallel CA$,therefore $\triangle ABC \sim \triangle PBR$.
And $\frac{PB}{AB} = \frac{RP}{CA} = \frac{BR}{BC} = \frac{4}{3}$.
Hence,the new triangle is similar to the given triangle with sides $\frac{4}{3}$ times the corresponding sides of the isosceles $\triangle ABC$.

(N/A) Steps of construction:
$1.$ Draw a line segment $AB = 5 \, cm$.
$2.$ At point $B$,construct $\angle ABY = 60^{\circ}$ and cut off $BC = 6 \, cm$ on $BY$.
$3.$ Join $AC$. Thus,$\triangle ABC$ is the required triangle.
$4.$ Draw any ray $AX$ making an acute angle with $AB$ on the side opposite to vertex $C$.
$5.$ Locate $7$ points $B_1, B_2, B_3, B_4, B_5, B_6, B_7$ on $AX$ such that $AB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.
$6.$ Join $B_7B$. Through $B_5$,draw a line parallel to $B_7B$ intersecting $AB$ at $M$.
$7.$ Through $M$,draw a line parallel to $BC$ intersecting $AC$ at $N$.
Then,$\triangle AMN$ is the required triangle whose sides are $\frac{5}{7}$ of the corresponding sides of $\triangle ABC$.
Justification:
Since $MN \parallel BC$,by Basic Proportionality Theorem,$\triangle AMN \sim \triangle ABC$.
By construction,$B_5M \parallel B_7B$. In $\triangle ABB_7$,by Thales theorem:
$\frac{AM}{AB} = \frac{AB_5}{AB_7} = \frac{5}{7}$.
Since $\triangle AMN \sim \triangle ABC$,the ratio of their corresponding sides is $\frac{AM}{AB} = \frac{AN}{AC} = \frac{MN}{BC} = \frac{5}{7}$.

(N/A) Steps of construction:
$1.$ Draw a circle with centre $O$ and radius $4 \, cm$.
$2.$ Draw a radius $OA$ and extend it to $B$ such that $OA = AB = 4 \, cm$. Thus,$OB = 8 \, cm$.
$3.$ Draw a circle with centre $A$ and radius $AO = 4 \, cm$. Let this circle intersect the original circle at points $P$ and $Q$.
$4.$ Join $BP$ and $BQ$. $BP$ and $BQ$ are the required tangents.
Justification:
In $\triangle OAP$,$OA = OP = 4 \, cm$ (radii) and $AP = 4 \, cm$ (radius of circle with centre $A$).
Since all sides are equal,$\triangle OAP$ is an equilateral triangle.
Therefore,$\angle OAP = 60^{\circ}$.
Since $OAB$ is a straight line,$\angle BAP = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
In $\triangle BAP$,$BA = AP = 4 \, cm$. Thus,$\triangle BAP$ is an isosceles triangle.
Therefore,$\angle ABP = \angle APB = (180^{\circ} - 120^{\circ}) / 2 = 30^{\circ}$.
Similarly,$\angle ABQ = 30^{\circ}$.
Thus,$\angle PBQ = \angle ABP + \angle ABQ = 30^{\circ} + 30^{\circ} = 60^{\circ}$.
Measurement:
The distance between the centre $O$ and the point of intersection $B$ is $OB = OA + AB = 4 \, cm + 4 \, cm = 8 \, cm$.

(N/A) Steps of construction:
$1.$ Draw a line segment $BC = 6 \, cm$.
$2.$ Taking $B$ and $C$ as centers,draw two arcs of radii $4 \, cm$ and $9 \, cm$ respectively,intersecting each other at $A$.
$3.$ Join $BA$ and $CA$. $\triangle ABC$ is the required triangle.
$4.$ From $B$,draw any ray $BX$ downwards making an acute angle.
$5.$ Mark three points $B_1, B_2, B_3$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3$.
$6.$ Join $B_2C$ and from $B_3$,draw $B_3M \parallel B_2C$ intersecting the extended line segment $BC$ at $M$.
$7.$ From point $M$,draw $MN \parallel CA$ intersecting the extended line segment $BA$ at $N$.
Then,$\triangle NBM$ is the required triangle whose sides are $\frac{3}{2}$ times the corresponding sides of $\triangle ABC$.
The two triangles are not congruent because congruent triangles must have the same shape and size. Here,the triangles are similar (same shape),but their sizes are different.
Justification:
Since $B_3M \parallel B_2C$,by Basic Proportionality Theorem:
$\frac{BC}{CM} = \frac{BB_2}{B_2B_3} = \frac{2}{1}$.
Now,$\frac{BM}{BC} = \frac{BC + CM}{BC} = 1 + \frac{CM}{BC} = 1 + \frac{1}{2} = \frac{3}{2}$.
Also,$MN \parallel CA$,therefore $\triangle ABC \sim \triangle NBM$.
Thus,$\frac{NB}{AB} = \frac{NM}{AC} = \frac{BM}{BC} = \frac{3}{2}$.

(N/A) Data: $\overline{ AB }$ of length $5.9\,cm$ is given.
To construct: $\overline{ AB }$ is to be divided in the ratio $5: 7$ from $A$.
Steps of construction:
$(1)$ In different half-planes of $\overleftrightarrow{ AB }$ containing $\overline{ AB }$,draw $\overrightarrow{ AX }$ and $\overrightarrow{ BY }$ such that $\angle XAB$ and $\angle YBA$ are congruent acute angles.
$(2)$ With some appropriate radius and centre $A$,draw an arc to intersect $\overrightarrow{ AX }$ at $A_{1}$. Similarly,with centre $A_{1}$ and the same radius,draw an arc to intersect $\overrightarrow{ AX }$ at $A_{2}$ such that $A-A_{1}-A_{2}$. Similarly,draw an arc with the same radius and centre $A_{k}$ to intersect $\overrightarrow{ AX }$ at $A_{k+1}$ such that $A_{k-1}-A_{k}-A_{k+1}$,where $k=2, 3, 4$. Thus,we obtain five points $A_{1}, A_{2}, A_{3}, A_{4}$ and $A_{5}$ on $\overrightarrow{ AX }$ such that $AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}$.
$(3)$ Now,with the same radius and beginning with point $B$ as centre,obtain seven points $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$ and $B_{7}$ on $\overrightarrow{ BY }$ such that $BB_{1} = B_{1}B_{2} = \dots = B_{6}B_{7}$.
$(4)$ Draw $\overline{A_{5}B_{7}}$ to intersect $\overline{ AB }$ at $M$.
Thus,$M \in \overline{ AB }$ is the point which divides $\overline{ AB }$ in the ratio $5: 7$.
Justification: Here $\frac{AA_{5}}{BB_{7}} = \frac{5}{7}$. Since $\overrightarrow{AX} \parallel \overrightarrow{BY}$,the correspondence $AA_{5}M \leftrightarrow BB_{7}M$ between $\Delta AA_{5}M$ and $\Delta BB_{7}M$ is a similarity.
$\therefore \frac{AM}{BM} = \frac{AA_{5}}{BB_{7}} = \frac{5}{7}$.

(N/A) Data: $\overline{AB}$ is given.
To construct: $\overline{AB}$ is to be divided into three parts in the ratio $2: 3: 4$ from $A$.
Steps of construction:
$(1)$ In different half-planes of $\overleftrightarrow{AB}$ containing $\overline{AB}$,draw $\overrightarrow{AX}$ and $\overrightarrow{BY}$ such that $\angle XAB$ and $\angle YBA$ are congruent acute angles.
$(2)$ With some appropriate radius and centre $A$,draw an arc to intersect $\overrightarrow{AX}$ at $A_1$. Similarly,with centre $A_1$ and the same radius,draw an arc to intersect $\overrightarrow{AX}$ at $A_2$ such that $A-A_1-A_2$. Similarly,draw an arc with the same radius and centre $A_k$ to intersect $\overrightarrow{AX}$ at $A_{k+1}$ such that $A_{k-1}-A_k-A_{k+1}$,where $k=2, 3, 4, \dots, 8$. Thus,we obtain nine points $A_1, A_2, \dots, A_9$ on $\overrightarrow{AX}$ such that $AA_1 = A_1A_2 = \dots = A_8A_9$.
$(3)$ Now,with the same radius and beginning with point $B$ as centre,obtain nine points $B_1, B_2, \dots, B_9$ on $\overrightarrow{BY}$ such that $BB_1 = B_1B_2 = \dots = B_8B_9$.
$(4)$ Draw $\overline{AB_9}$,$\overline{A_2B_7}$,$\overline{A_5B_4}$,and $\overline{A_9B}$ so that $\overline{A_2B_7}$ intersects $\overline{AB}$ at $M$ and $\overline{A_5B_4}$ intersects $\overline{AB}$ at $N$.
Thus,we obtain points $M$ and $N$ which divide $\overline{AB}$ in the ratio $2: 3: 4$ from $A$,i.e.,$AM : MN : NB = 2: 3: 4$.
Justification: Here,the intercepts made on transversals $\overleftrightarrow{AX}$,$\overleftrightarrow{AB}$,and $\overleftrightarrow{BY}$ by $\overline{A_2B_7} \parallel \overline{A_5B_4}$ are proportionate.
Hence,$AM : MN : NB = AA_2 : A_2A_5 : A_5A_9 = B_9B_7 : B_7B_4 : B_4B = 2: 3: 4$.

(N/A) Data: Construct $\Delta ABC$ in which $AB = 4 \, cm$,$BC = 7 \, cm$,and $AC = 5 \, cm$.
To construct: Construct $\Delta BPQ$ similar to $\Delta ABC$ such that the ratio of their corresponding sides is $2:3$.
Steps of construction:
$(1)$ Construct $\Delta ABC$ in which $AB = 4 \, cm$,$BC = 7 \, cm$,and $AC = 5 \, cm$.
$(2)$ In the half-plane of $\overleftrightarrow{BC}$ that does not contain $A$,draw a ray $\overrightarrow{BX}$ such that $\angle CBX$ is an acute angle.
$(3)$ With an appropriate radius and center $B$,draw an arc to intersect $\overrightarrow{BX}$ at $B_1$. Similarly,with center $B_1$ and the same radius,draw an arc to intersect $\overrightarrow{BX}$ at $B_2$ such that $B-B_1-B_2$. Again,with the same radius and center $B_2$,draw an arc to intersect $\overrightarrow{BX}$ at $B_3$ such that $B_1-B_2-B_3$.
$(4)$ Draw the line segment $\overline{B_3C}$.
$(5)$ Through $B_2$,draw a line parallel to $\overline{B_3C}$ to intersect $\overline{BC}$ at $P$.
$(6)$ Through $P$,draw a line parallel to $\overline{CA}$ to intersect $\overline{AB}$ at $Q$.
Thus,$\Delta BPQ$ is the required triangle.
Justification: $\overleftrightarrow{BX}$ and $\overleftrightarrow{BC}$ are transversals to $\overleftrightarrow{B_3C} \parallel \overleftrightarrow{B_2P}$.
Therefore,$BP:BC = BB_2:BB_3 = 2:3$.
Similarly,$\overleftrightarrow{BC}$ and $\overleftrightarrow{BA}$ are transversals to $\overleftrightarrow{CA} \parallel \overleftrightarrow{PQ}$.
Therefore,$BQ:BA = BP:BC = 2:3$.

(N/A) Data: Construct $\Delta PQR$ with $m \angle P = 60^{\circ}, m \angle Q = 45^{\circ}$ and $PQ = 6 \text{ cm}$.
To construct: Construct $\Delta PBC$ similar to $\Delta PQR$ such that the ratio of their corresponding sides is $5:3$.
Steps of construction:
$(1)$ Construct $\Delta PQR$ with $m \angle P=60^{\circ}, m \angle Q=45^{\circ}$ and $PQ = 6 \text{ cm}$.
$(2)$ In the half-plane of $\overleftrightarrow{PQ}$ that does not contain $R$,draw a ray $\overrightarrow{PZ}$ such that $\angle QPZ$ is an acute angle.
$(3)$ With some appropriate radius and center $P$,draw an arc to intersect $\overrightarrow{PZ}$ at $P_1$. With the same radius and center $P_1$,draw an arc to intersect $\overrightarrow{PZ}$ at $P_2$ such that $P-P_1-P_2$. Similarly,with the same radius and center $P_k$,draw an arc to intersect $\overrightarrow{PZ}$ at $P_{k+1}$ such that $P_{k-1}-P_k-P_{k+1}$,where $k=2, 3, 4$.
$(4)$ Draw $\overline{P_3Q}$.
$(5)$ Through $P_5$,draw a ray parallel to $\overline{P_3Q}$ to intersect the extended ray $\overrightarrow{PQ}$ at $B$.
$(6)$ Through $B$,draw a ray parallel to $\overline{QR}$ to intersect the extended ray $\overrightarrow{PR}$ at $C$.
Thus,$\Delta PBC$ is the required triangle.
Justification: In $\Delta PP_5B$,$P-P_3-P_5$,$P-Q-B$ and $\overline{P_3Q} \parallel \overline{P_5B}$.
$\therefore \frac{PB}{PQ} = \frac{PP_5}{PP_3} = \frac{5}{3}$.
Similarly,in $\Delta PBC$,$P-Q-B$,$P-R-C$ and $\overline{QR} \parallel \overline{BC}$.
$\therefore \frac{PC}{PR} = \frac{PB}{PQ} = \frac{5}{3}$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{PQ} = 8.2 \, cm$ using a ruler.
$2$. Draw an acute angle $\angle QPX$ at point $P$ below the line segment $\overline{PQ}$.
$3$. Starting from $P$,mark $3 + 7 = 10$ points $A_1, A_2, A_3, \dots, A_{10}$ on the ray $PX$ such that $PA_1 = A_1A_2 = A_2A_3 = \dots = A_9A_{10}$.
$4$. Join the point $A_{10}$ to $Q$.
$5$. Through point $A_3$,draw a line parallel to $A_{10}Q$ intersecting $\overline{PQ}$ at point $R$.
$6$. The point $R$ divides $\overline{PQ}$ in the ratio $3:7$ such that $PR:RQ = 3:7$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{AB}$ of length $10 \,cm$ using a ruler.
$2$. Draw any ray $AX$ making an acute angle with $\overline{AB}$.
$3$. Locate $3 + 8 = 11$ points $A_1, A_2, ..., A_{11}$ on ray $AX$ such that $AA_1 = A_1A_2 = ... = A_{10}A_{11}$.
$4$. Join $B$ to $A_{11}$.
$5$. Through the point $A_3$,draw a line parallel to $A_{11}B$ (by making an angle equal to $\angle AA_{11}B$) intersecting $\overline{AB}$ at point $P$.
$6$. Thus,$P$ is the point dividing $\overline{AB}$ in the ratio $3:8$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{XY}$ of length $9\, cm$ using a ruler.
$2$. Draw a ray $XP$ making an acute angle with $\overline{XY}$ at point $X$.
$3$. Locate $2 + 5 = 7$ points $A_1, A_2, A_3, A_4, A_5, A_6, A_7$ on ray $XP$ such that $XA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7$.
$4$. Join $A_7$ to $Y$.
$5$. Through point $A_2$,draw a line parallel to $A_7Y$ (by making corresponding angles equal) intersecting $\overline{XY}$ at point $Z$.
$6$. Thus,point $Z$ divides $\overline{XY}$ in the ratio $2:5$.

(N/A) Steps of construction:
$1$. Draw a line segment $AB$ of length $7 \,cm$ using a ruler.
$2$. Draw a ray $AX$ making an acute angle with $AB$.
$3$. Locate $3 + 5 = 8$ points $A_1, A_2, A_3, A_4, A_5, A_6, A_7, A_8$ on ray $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7 = A_7A_8$.
$4$. Join the point $A_8$ to $B$.
$5$. Through the point $A_3$,draw a line parallel to $A_8B$ (by making corresponding angles equal) to intersect $AB$ at point $M$.
$6$. Thus,$M$ is the point on $AB$ which divides it in the ratio $3 : 5$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{XY} = 7.5\, cm$ using a ruler.
$2$. Draw an acute angle $\angle YXA$ at point $X$ below the segment $\overline{XY}$.
$3$. Starting from $X$,mark $3 + 4 + 5 = 12$ points $X_1, X_2, ..., X_{12}$ on the ray $XA$ such that $XX_1 = X_1X_2 = ... = X_{11}X_{12}$ using a compass.
$4$. Join the point $X_{12}$ to $Y$.
$5$. Through points $X_3$ and $X_7$ (since $3$ and $3+4=7$),draw lines parallel to $X_{12}Y$ intersecting $\overline{XY}$ at points $P$ and $Q$ respectively.
$6$. The points $P$ and $Q$ divide the segment $\overline{XY}$ in the ratio $3:4:5$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{PQ}$ of length $8.5 \text{ cm}$ using a ruler.
$2$. Draw any ray $PY$ making an acute angle with $\overline{PQ}$.
$3$. Locate $4 + 7 = 11$ points $A_1, A_2, \dots, A_{11}$ on ray $PY$ such that $PA_1 = A_1A_2 = A_2A_3 = \dots = A_{10}A_{11}$.
$4$. Join $A_{11}$ to $Q$.
$5$. Through the point $A_4$ (since the ratio is $4:7$),draw a line parallel to $A_{11}Q$ intersecting $\overline{PQ}$ at point $X$.
$6$. Thus,$X$ is the point on $\overline{PQ}$ which divides it in the ratio $4 : 7$.

(N/A) Steps of construction:
$1$. Draw a line segment $\overline{ AB }$ of length $7 \,cm$ using a ruler.
$2$. Draw an acute angle $\angle BAX$ at point $A$ with the line segment $\overline{ AB }$.
$3$. Mark $5$ points $A_1, A_2, A_3, A_4, A_5$ on the ray $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5$ using a compass.
$4$. Join $A_5$ to $B$.
$5$. Through point $A_2$,draw a line parallel to $A_5B$ intersecting $\overline{ AB }$ at point $M$.
$6$. Thus,$M$ is the required point on $\overline{ AB }$ such that $AM : AB = 2 : 5$.

(N/A) Step $1$: Draw a line segment $\overline{AB} = 8\, cm$ using a ruler.
Step $2$: Draw an acute angle $\angle BAX$ at point $A$ with the line segment $\overline{AB}$.
Step $3$: Mark $2 + 3 + 5 = 10$ points $A_1, A_2, A_3, \dots, A_{10}$ on the ray $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = \dots = A_9A_{10}$ using a compass.
Step $4$: Join the last point $A_{10}$ to point $B$.
Step $5$: Through points $A_2$ and $A_5$ (since $2$ and $2+3=5$),draw lines parallel to $A_{10}B$ intersecting $\overline{AB}$ at points $P$ and $Q$ respectively.
Step $6$: The points $P$ and $Q$ divide the line segment $\overline{AB}$ in the ratio $2:3:5$.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 6 \, cm$.
$2$. With $B$ as the center and radius $3 \, cm$,draw an arc. With $C$ as the center and radius $5 \, cm$,draw another arc intersecting the first arc at $A$.
$3$. Join $AB$ and $AC$ to obtain $\Delta ABC$.
$4$. Draw a ray $BX$ making an acute angle with $BC$ on the side opposite to vertex $A$.
$5$. Locate $4$ points $B_1, B_2, B_3, B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$6$. Join $B_4C$.
$7$. Draw a line through $B_3$ parallel to $B_4C$ intersecting $BC$ at $Q$.
$8$. Draw a line through $Q$ parallel to $AC$ intersecting $AB$ at $P$.
$9$. $\Delta BPQ$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 7 \text{ cm}$.
$2$. With $B$ as the center and radius $6 \text{ cm}$,draw an arc. With $C$ as the center and radius $5 \text{ cm}$,draw another arc intersecting the first arc at $A$.
$3$. Join $AB$ and $AC$ to form $\Delta ABC$.
$4$. Draw a ray $BX$ making an acute angle with $BC$ on the side opposite to vertex $A$.
$5$. Locate $5$ points $B_1, B_2, B_3, B_4, B_5$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.
$6$. Join $B_5C$.
$7$. From $B_4$,draw $B_4Y \parallel B_5C$ meeting $BC$ at $Y$.
$8$. From $Y$,draw $YX \parallel CA$ meeting $AB$ at $X$. Then $\Delta BXY$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $QR = 7 \, cm$.
$2$. At point $Q$,construct an angle of $45^{\circ}$ using a protractor or compass.
$3$. At point $R$,construct an angle of $60^{\circ}$ using a compass.
$4$. Let the intersection of these two rays be point $P$. Thus,$\Delta PQR$ is formed.
$5$. Draw an acute angle $\angle RQX$ below the base $QR$.
$6$. Locate $4$ points $Q_1, Q_2, Q_3, Q_4$ on ray $QX$ such that $QQ_1 = Q_1Q_2 = Q_2Q_3 = Q_3Q_4$.
$7$. Join $Q_3$ to $R$. Draw a line through $Q_4$ parallel to $Q_3R$ intersecting the extended line segment $QR$ at $R'$.
$8$. Draw a line through $R'$ parallel to $RP$ intersecting the extended line segment $QP$ at $P'$.
$9$. $\Delta P'QR'$ is the required triangle similar to $\Delta PQR$ with a scale factor of $\frac{4}{3}$.

(N/A) Steps of construction:
$1$. Draw an equilateral triangle $\Delta XYZ$ with each side equal to $6\, cm$.
$2$. Draw a ray $XS$ making an acute angle with $XY$ at vertex $X$ on the side opposite to $Z$.
$3$. Locate $5$ points $X_1, X_2, X_3, X_4, X_5$ on ray $XS$ such that $XX_1 = X_1X_2 = X_2X_3 = X_3X_4 = X_4X_5$.
$4$. Join $X_4$ to $Y$. Draw a line through $X_5$ parallel to $X_4Y$ intersecting the extended line segment $XY$ at point $M$.
$5$. Draw a line through $M$ parallel to $YZ$ intersecting the extended line segment $XZ$ at point $N$.
$6$. Thus,$\Delta XMN$ is the required triangle similar to $\Delta XYZ$ with scale factor $\frac{5}{4}$.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 6 \text{ cm}$.
$2$. At point $B$,construct an angle of $90^{\circ}$ using a protractor or compass.
$3$. Cut an arc of $5 \text{ cm}$ on the ray from $B$ to mark point $A$. Join $AC$ to complete $\Delta ABC$.
$4$. Draw an acute angle $\angle CBX$ below $BC$.
$5$. Mark $3$ points $B_1, B_2, B_3$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3$.
$6$. Join $B_2$ to $C$. Draw a line through $B_3$ parallel to $B_2C$ intersecting the extended line $BC$ at $Q$.
$7$. Draw a line through $Q$ parallel to $AC$ intersecting the extended line $BA$ at $P$.
$8$. $\Delta BPQ$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 8 \, cm$.
$2$. At point $B$,construct an angle of $60^\circ$ using a compass and ruler.
$3$. From point $B$,cut an arc of $6 \, cm$ on the ray making $60^\circ$ to mark point $A$. Join $AC$ to complete $\Delta ABC$.
$4$. Draw an acute angle $\angle CBZ$ below the line segment $BC$.
$5$. Mark $5$ points $B_1, B_2, B_3, B_4, B_5$ on ray $BZ$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.
$6$. Join $B_5$ to $C$.
$7$. Draw a line from $B_4$ parallel to $B_5C$ intersecting $BC$ at $Y$.
$8$. Draw a line from $Y$ parallel to $AC$ intersecting $AB$ at $X$.
$9$. $\Delta BXY$ is the required triangle similar to $\Delta ABC$ with a scale factor of $\frac{4}{5}$.

(N/A) Steps of construction:
$1$. Draw a line segment $YZ = 6\, cm$.
$2$. With $Y$ as the center and radius $4\, cm$,draw an arc.
$3$. With $Z$ as the center and radius $7\, cm$,draw another arc intersecting the previous arc at point $X$.
$4$. Join $XY$ and $XZ$ to form $\Delta XYZ$.
$5$. Draw an acute angle $\angle ZYA$ at vertex $Y$ below the line $YZ$.
$6$. Mark $4$ points $Y_1, Y_2, Y_3, Y_4$ on $YA$ such that $YY_1 = Y_1Y_2 = Y_2Y_3 = Y_3Y_4$.
$7$. Join $Y_3$ to $Z$.
$8$. Draw a line through $Y_4$ parallel to $Y_3Z$ intersecting the extended line $YZ$ at $N$.
$9$. Draw a line through $N$ parallel to $XZ$ intersecting the extended line $YX$ at $M$.
$10$. $\Delta YMN$ is the required triangle.

(N/A) Data: Draw a circle with centre $A$ and radius $5 \, cm$. Take a point $B$ in its exterior such that $AB = 8 \, cm$.
To construct: Through point $B$,draw two tangents to the circle with centre $A$ and radius $5 \, cm$.
Steps of construction:
$1$. Draw a circle with centre $A$ and radius $5 \, cm$ and mark a point $B$ outside the circle such that $AB = 8 \, cm$.
$2$. Draw the line segment $\overline{AB}$.
$3$. Find the midpoint $M$ of $\overline{AB}$ by constructing its perpendicular bisector.
$4$. With $M$ as the centre and $MA$ as the radius,draw a circle that intersects the original circle at points $X$ and $Y$.
$5$. Draw the rays $\overrightarrow{BX}$ and $\overrightarrow{BY}$.
Thus,$\overleftrightarrow{BX}$ and $\overleftrightarrow{BY}$ are the required tangents.
Calculation: In $\triangle AXB$,$\angle AXB = 90^\circ$ (angle in a semicircle). By Pythagoras theorem,$AB^2 = AX^2 + BX^2$.
$8^2 = 5^2 + BX^2 \implies 64 = 25 + BX^2 \implies BX^2 = 39 \implies BX = \sqrt{39} \approx 6.24 \, cm$.
Therefore,the length of the tangents is approximately $6.24 \, cm$.

(N/A) Data: Draw a circle with the help of a circular bangle and take point $A$ in the exterior of the circle.
To construct: Through $A$,tangents are to be drawn to the circle.
$(1)$ Draw a circle with the help of a circular bangle and take point $A$ in the exterior of the circle.
$(2)$ Draw two non-parallel chords $\overline{PQ}$ and $\overline{RS}$ in this circle.
$(3)$ Draw the perpendicular bisectors of $\overline{PQ}$ and $\overline{RS}$ to intersect each other at $O$. Here,$O$ is the centre of the circle drawn with the help of the circular bangle.
$(4)$ Draw $\overline{OA}$.
$(5)$ Obtain the midpoint $M$ of $\overline{OA}$ by constructing the perpendicular bisector of $\overline{OA}$.
$(6)$ Draw a circle with centre $M$ and radius $MA$ to intersect the first circle at $X$ and $Y$.
$(7)$ Draw rays $\overrightarrow{AX}$ and $\overrightarrow{AY}$.
Thus,$\overleftrightarrow{AX}$ and $\overleftrightarrow{AY}$ are the required tangents.

(N/A) Data: Draw $\overline{AB}$ such that $AB = 10 \, cm$. Draw $\odot(A, 3 \, cm)$ and $\odot(B, 4 \, cm)$.
To construct: From the centre of each circle,tangents are to be drawn to the other circle.
Steps of construction:
$(1)$ Draw $\overline{AB}$ such that $AB = 10 \, cm$. Draw $\odot(A, 3 \, cm)$ and $\odot(B, 4 \, cm)$.
$(2)$ Obtain the midpoint $M$ of $\overline{AB}$ by constructing its perpendicular bisector.
$(3)$ Draw $\odot(M, MA)$ to intersect $\odot(A, 3 \, cm)$ at $S$ and $T$,and $\odot(B, 4 \, cm)$ at $X$ and $Y$.
$(4)$ Draw $\overrightarrow{AX}$ and $\overrightarrow{AY}$ as well as $\overrightarrow{BS}$ and $\overrightarrow{BT}$.
Thus,$\overleftrightarrow{AX}$ and $\overleftrightarrow{AY}$ are the tangents from $A$ to $\odot(B, 4 \, cm)$ and $\overleftrightarrow{BS}$ and $\overleftrightarrow{BT}$ are the tangents from $B$ to $\odot(A, 3 \, cm)$.

(N/A) Data: Draw $\odot(P, 4 \text{ cm})$.
To construct: Draw a pair of tangents to $\odot(P, 4 \text{ cm})$ such that the measure of the angle between the tangents at their point of intersection $A$ is $60^\circ$.
$(1)$ Draw $\odot(P, 4 \text{ cm})$ and two radii $\overline{PR}$ and $\overline{PQ}$ such that $m\angle RPQ = 180^\circ - 60^\circ = 120^\circ$.
$(2)$ Through $R$,draw a line perpendicular to $\overline{PR}$.
$(3)$ Through $Q$,draw a line perpendicular to $\overline{PQ}$.
$(4)$ Let the lines drawn in step $(2)$ and step $(3)$ intersect at $A$.
Thus,$\overleftrightarrow{AR}$ and $\overleftrightarrow{AQ}$ are the required tangents such that the measure of the angle between them is $60^\circ$.

(N/A) Steps of construction:
$1$. Draw a circle with center $P$ and radius $3\, cm$.
$2$. Mark a point $R$ outside the circle such that the distance $PR = 6\, cm$.
$3$. Find the midpoint $M$ of the line segment $PR$ by drawing its perpendicular bisector.
$4$. With $M$ as the center and $MP$ (or $MR$) as the radius,draw a second circle.
$5$. Let this circle intersect the original circle at points $A$ and $B$.
$6$. Join $RA$ and $RB$. These are the required tangents to the circle $\odot(P, 3\, cm)$ from point $R$.