NCERT Solutions Ganita Prakash Chapter 2 – 21Section 2.5 Angle — Figure it Out
Book page 19 Updated on2026-09-05
Q1.
Can you find the angles in the given pictures? Draw the rays forming any one of the angles and name the vertex of the angle.
Answer
Yes. In the picture of the open scissors-like shape, one clear angle is ∠BDC.
Vertex: D — the point where the two edges meet.
Arms: the rays DB and DC drawn along the two edges.
An angle: two rays BD and BE from the common starting point B. B is the vertex; BD and BE are the arms.
Try This: look for angles in the other pictures too — the open beak of a bird, the hands of a clock, the corner of a table, the branches of a tree. In each case first find the point where the two edges meet (the vertex), then draw the two rays along the edges.
Q2.
Draw and label an angle with arms ST and SR.
Answer
The two arms are ST and SR, so both rays must start from the same point S — that makes S the vertex.
The angle ∠TSR — both arms ST and SR start from the same point S, which is therefore the vertex.
Steps:
Mark a point and label it S.
From S draw a ray and mark a point T on it.
From the same point S draw another ray in a different direction and mark a point R on it.
Draw a small curve near S between the arms and name the angle ∠TSR (or ∠RST).
Tip: the vertex letter always sits in the middle of the name — that is why it is ∠TSR and not ∠STR.
Q3.
Explain why ∠APB cannot be labelled as ∠P.
Answer
Because at the point P there is more than one angle, so the single letter P would not tell us which one is meant.
Three rays from P make three different angles — so the name “∠P” would be ambiguous.
In the figure, three rays PA, PB and PC all start at P. That gives three different angles at the same vertex:
∠APB ∠BPC ∠APC
Why the rule exists: the name ∠P only fixes the vertex; it does not fix the two arms. Using a point on each arm — A on one arm and B on the other — makes the angle unmistakable. We may write ∠P only when exactly one angle is formed at P.
Q4.
Name the angles marked in the given figure.
Answer
The two marked angles at the vertex T. Measured on the page, ∠PTR ≈ 104° and ∠QTR ≈ 64°.
The two marked angles are
∠PTR (or ∠RTP) and ∠QTR (or ∠RTQ)
Both have their vertex at T. The first is formed by the arms TP and TR, the second by the arms TQ and TR.
Tip: the vertex T is written in the middle of each name. If you measure the printed figure you will find ∠PTR ≈ 104° (obtuse) and ∠QTR ≈ 64° (acute).
Q5.
Mark any three points on your paper that are not on one line. Label them A, B, C. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C? Write them down, and mark each of them with a curve as in Fig. 2.9.
Answer
Three points not on a line give 3 lines and 3 angles, one at each corner.
Number of lines: 3. They are AB, BC and CA.
Pairs of points: (A, B), (B, C), (C, A) → 3 lines
Number of angles: 3. One at each corner:
∠ABC (or ∠CBA) ∠BCA (or ∠ACB) ∠CAB (or ∠BAC)
Why 3 and not more: at each point exactly two lines meet, and two rays make just one angle. Three corners × 1 angle = 3 angles. The three points must not lie on one line — if they did, no angle would be formed at all (only a straight angle).
Q6.
Now mark any four points on your paper so that no three of them are on one line. Label them A, B, C, D. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C, D? Write them all down, and mark each of them with a curve as in Fig. 2.9.
Answer
Four points, no three on a line, give 6 lines: AB, BC, CD, DA, AC and BD.
Number of lines: 6.
AB, BC, CD, DA, AC, BD
Number of angles: 12 — three at each of the four points.
Vertex
Rays from it
Angles
A
AB, AC, AD
∠BAC, ∠CAD, ∠BAD
B
BA, BC, BD
∠ABC, ∠CBD, ∠ABD
C
CA, CB, CD
∠ACB, ∠BCD, ∠ACD
D
DA, DB, DC
∠ADB, ∠BDC, ∠ADC
Why the count works out: with 4 points you can pick a pair in 4 × 3 ÷ 2 = 6 ways, so there are 6 lines. At each point three lines meet; choosing 2 of those 3 rays gives 3 × 2 ÷ 2 = 3 angles. Four vertices × 3 = 12 angles.