# Ex.11.1 Q3 Constructions Solution - NCERT Maths Class 9

## Question

Construct the angles of the following measurements:

(i) \( 30^{\circ} \)

(ii) \(22 \frac{1}{2}^{\circ}\)

(iii) \(15^{\circ}\)

## Text Solution

\({\rm{(i)}}\;\;30^{\circ}\)

**Reasoning:**

We need to construct an angle of \(60\) degrees and then bisect it to get an angle measuring \(30^\circ\)

**Steps of Construction:**

(i) Draw ray \(PQ\).

(ii) To construct an angle of \(60^{\circ}\) .

With \(P\) as centre and any radius, draw a wide arc to intersect \(PQ\) at \(R\). With \(R\) as centre and same radius draw an arc to intersect the initial arc at \(S\). \(\angle {SPR}=60^{\circ}\)

(iii) To bisect \(\angle {SPR}\)

With \(R\) and \(S\) as centres and same radius draw two arcs to intersect at \(T\). Join \(P\) and \(T\) i.e. \(PT\) is the angle bisector. Hence,

\[\begin{align}\angle {TPR}=\frac{1}{2} \angle {SPR}=30^{\circ}\end{align}\]

\({\rm{(ii)}}\;\;22 \frac{1}{2}^{\circ}\)

**Reasoning:**

We need to construct two adjacent angles of and bisect the second one to get a angle. This has to be bisected again to get a \(45^\circ\) angle. The \(45^\circ\) angle has to be further bisected to get \(\begin{align}22 \frac{1}{2}^{\circ}\end{align}\) angle.

\[\begin{align} 22 \frac{1}{2}^{\circ} &=\frac{45^{\circ}}{2} \\ 45^{\circ} &=\frac{90^{\circ}}{2}=\frac{30^{\circ}+60^{\circ}}{2} \end{align}\]

**Steps of Construction:**

(i) Draw ray \(PQ\)

(ii) To construct an angle of \(60^{\circ}\)

With \(P\) as center and any radius draw a wide arc to intersect \(PQ\) at \(R\). With \(R\) as center and same radius draw an arc to intersect the initial arc at \(S\). \(\angle {SPR}=60^{\circ}\)

(iii)To construct adjacent angle of \(60^{\circ}\) **.**

With \(S\) as the center and same radius as before, draw an arc to intersect the initial arc at \(T\) \(\angle {TPS}=60^{\circ}\) .

(iv)To bisect \(\angle {TPS}\)

With \(T\) and \(S\) as centers and same radius as before, draw arcs to intersect each other at \(Z\) Join \(P\) and \(Z\) \(\angle {ZPQ}=90^{\circ}\)

(v) To bisect \(\angle {ZPQ}\)

With \(R\) and \(U\) as centers and radius than half of \(RU\), draw arcs to intersect each other at \(V\). Join \(P\) and \(V\). \(\angle {VPQ}=45^{0}\)

(vi) To bisect \(\angle {VPQ}=45^{0}\)

With \(W\) and \(R\) as centers and radius greater than half of \(WR\), draw arcs to intersect each other at \(X\). Join \(P\) and \(X\). \(PX\) bisects \(\angle {VPQ}\)

Hence,

\[\begin{align} \angle {XPQ} &=\frac{1}{2} \angle {WPQ} \\ &=\frac{1}{2} \times 45^{0} \\ &=22 \frac{1}{2} \end{align}\]

\({\rm{(iii)}}\;\;15^{\circ}\)

**Reasoning:**

We need to construct an angle of 60 degrees and then bisect it to get an angle measuring \(30^\circ\). This has to be bisected again to get a \(15^\circ\) angle.

\[\begin{align}15^{0}=\frac{30^{\circ}}{2}=\frac{\frac{60^{0}}{2}}{2}\end{align}\]

**Steps of Construction:**

(i) Draw ray \(PQ\).

(ii) To construct an angle of \(60^{\circ}\) .

With \(P\) as center and any radius draw a wide arc to intersect \(PQ\) at \(R\). With \(R\) as center and same radius draw an arc to intersect the initial arc at \(S\). \(\angle {SPR}=60^{\circ}\)

(iii) Bisect \(\angle {SPR}\) . With \(R\) and \(S\) as centers and radius greater than half of \(RS\) draw arcs to intersect each other at \(T\). Join \(P\) and \(T\) i.e. \(PT\) is the angle bisector of \(\angle {SPR}\) .

\[\begin{align} \angle {TPQ} &=\frac{1}{2} \angle{SPR} \\ &=\frac{1}{2} \times 60^{\circ} \\ &=30^{\circ} \end{align}\]

(iv)To bisect \(\angle {TPQ}\)

With \(R\) and \(W\) as centers and radius greater than half of \(RT\), draw arcs to intersect each other at \(U\) Join \(P\) and \(U\). \(PU\) is the angle bisector of \(\angle {TPQ}\) **.**

\[\begin{align}\angle {UPQ}=\frac{1}{2} \angle {TPQ}=15^{\circ}\end{align}\]