NCERT Solutions Ganita Prakash (Part 1) Chapter 5 –125Section 5.8 Alternate Angles — Figure it Out

Book page 123 Updated on2026-09-05

Q1.

Find the angles marked below. (Fig. 5.30 — a, b, c, d, e, f, g, h, i and j)

Answer

Each part is one step of corresponding, alternate or interior angles. In several parts an extra angle is printed which is not needed.

Angle	Value	Reason
a	48°	Slide the 48° angle down the transversal to the second parallel line — that corresponding angle is 48°, and a° is vertically opposite it
b	52°	b and 52° are alternate interior angles for the two parallel lines
c	81°	c and 99° are interior angles on the same side: c = 180° – 99°
d	99°	d and 81° are interior angles on the same side: d = 180° – 81°
e	69°	e and 69° are alternate interior angles (97° and 83° belong to the other transversal and are not needed)
f	48°	f and 132° are interior angles on the same side: f = 180° – 132°
g	122°	g and the 122° at the second vertical line are corresponding angles
h	75°	h and 75° are alternate interior angles (the 120° is not needed)
i	54°	The two arrow-marked rays are parallel, so the left base angle is 56°. In the triangle, i = 180° – 70° – 56°
j	97°	j and 97° are alternate interior angles for the first and third parallel lines cut by the same transversal (27° and 124° are not needed)

The two longer ones in full:

Part i: base angles of the triangle
Right base angle = 70°
Left base angle = 56° (corresponding to the 56° at the right foot, since the two rays are parallel)
i = 180° – 70° – 56° = 54°
Check at the right foot: 70° + 54° + 56° = 180° ✔

Part c: 99° + 81° = 180° at the upper line ✔
c + 99° = 180° → c = 81°

Why it happens: Whenever a transversal cuts parallel lines, only two angle sizes exist and they add to 180°. So every answer is either a copy of a printed angle or 180° minus it. Deciding which one is the whole job.

Tip: Before calculating, check for the arrow marks. If a pair of lines has no arrow marks, you cannot use these rules on that pair — that is exactly why parts e, h and j contain spare numbers.

Q2.

Find the angle represented by a. (Fig. 5.31 — four figures)

Answer

Figure	a	Working
(i) two parallel lines, 42° and 100° marked	138°	Corresponding angle of 42° at the lower line is 42°; a is its linear pair, so a = 180° – 42°
(ii) grid of two pairs of parallel lines, 62° marked	118°	62° carries down to the lower line, then across to the other slanting line; a is its linear pair, so a = 180° – 62°
(iii) three parallel lines, 110° and 35° marked	105°	110° carries down to the middle line; the 35° gap between the two slanting lines leaves 110° – 35° = 75°; a = 180° – 75°
(iv) two parallel slanting lines, 67° and a right angle	23°	In the right-angled triangle: a = 180° – 90° – 67°

(i) a = 180° – 42° = 138°
(ii) a = 180° – 62° = 118°
(iii) angle of the steep line with the middle line = 110° – 35° = 75°
a = 180° – 75° = 105°
(iv) a = 180° – 90° – 67° = 23°

Why it happens: In (iii) the two slanting lines start from the same point on the middle line, so the 35° between them can simply be subtracted from the 110° that was carried down from the top line. In (iv) the second slanting line is parallel to the first, so it also makes 67° with the base, and the three angles of the triangle must total 180°.

Check it yourself: In (i), the 100° angle belongs to a different transversal and is not needed — a good reminder to check which lines each printed angle sits between.

Q3.

In the figures below, what angles do x and y stand for? (Fig. 5.32)

Answer

First figure: x = 25°, y = 155°. Second figure: x = 25°.

First figure
The vertical line is perpendicular to the upper line, and the two horizontal lines are parallel
⟹ the vertical line is perpendicular to the lower line too
Angle between the vertical and the lower line = 90°
65° is the part of it taken by the slanting line
Remaining part = 90° – 65° = 25°
x = 25° (vertically opposite that part)

y is at the upper line, in the position matching x's linear pair
y = 180° – 25° = 155°

Second figure
At the lower line, the first transversal makes 53° and the second makes 78°
The two horizontal lines are parallel, so these angles are copied at the upper line
The two transversals meet on the upper line
x = 78° – 53° = 25°

Why it happens: In the second figure, x is the angle between the two transversals. Each transversal carries its own tilt from the lower line up to the upper line unchanged, so the angle between them is simply the difference of the two tilts.

Did you know? The angle between two transversals is the same wherever they are measured — this is exactly what makes the two figures give the same answer, 25°.

Q4.

In Fig. 5.33, ∠ABC = 45° and ∠IKJ = 78°. Find angles ∠GEH, ∠HEF, ∠FED

Answer

The two horizontal lines carry arrow marks, so they are parallel. Line J–K–E–F is one straight line and line C–B–E–H is another, and both pass through E.

∠GEH
∠ABC = 45° (given, at B on the upper line)
∠GEH is vertically opposite ∠DEB at E
∠DEB = ∠ABC = 45° (corresponding-type step for the parallel lines)
∠GEH = 45°

∠FED
∠IKJ = 78° (given, at K on the upper line)
∠AKE = 78° (vertically opposite ∠IKJ)
∠FED = 78° (the matching angle at E)
∠FED = 78°

∠HEF
∠GEH + ∠HEF + ∠FED = 180° (straight line GED)
45° + ∠HEF + 78° = 180°
∠HEF = 180° – 123° = 57°

So ∠GEH = 45°, ∠HEF = 57°, ∠FED = 78°.

Why it happens: The rays EH and EF are just the two given lines continued past E. Each line keeps its tilt all the way down, so the angle it made with the upper line reappears at the lower line. The three angles below E then fill the straight line GD, which is why they must add to 180°.

Check it yourself: 45 + 57 + 78 = 180 ✔ Also ∠BEK, the angle between the two lines above E, equals 57° — it is vertically opposite ∠HEF.

Q5.

In Fig. 5.34, AB is parallel to CD and CD is parallel to EF. Also, EA is perpendicular to AB. If ∠BEF = 55°, find the values of x and y.

Answer

x = 125° and y = 125°.

AB ∥ CD ∥ EF, and BE is a transversal of all three

∠BEF = 55° (given, at E between EB and EF)
At D: angle between DB and the downward part of CD = 55° (corresponding angles)
y and that angle form a linear pair along the transversal
y = 180° – 55° = 125°

At B: angle between the downward part of AB and BE
x = 180° – 55° = 125° (same reasoning at the third parallel line)

So x = y = 125°.

Why it happens: All three lines AB, CD and EF lean the same way, so the single transversal BE makes the same set of angles at every one of them. That is why x and y turn out to be equal, and equal to 180° – 55°.

Tip: "EA is perpendicular to AB" tells you that A, C and E lie on a line at right angles to the three parallels — a handy confirmation that the three lines really are parallel, though the value of x and y does not depend on it.

Q6.

What is the measure of angle ∠NOP in Fig. 5.35? [Hint: Draw lines parallel to LM and PQ through points N and O.]

Answer

∠NOP = 108°.

LM and PQ carry the same arrow mark, so LM ∥ PQ. Draw a dashed line through N parallel to LM, and another through O parallel to PQ. Now all four lines are parallel to one another.

The dashed violet lines through N and O are parallel to LM and PQ. They split ∠MNO into 40° + 56° and ∠NOP into 56° + 52°.

At N
∠LMN = 40° and LM ∥ (dashed line at N)
Angle between NM and the dashed line = 40° (alternate angles)
∠MNO = 96°
Angle between the dashed line and NO = 96° – 40° = 56°

At O
Dashed line at N ∥ dashed line at O
Angle between ON and the dashed line at O = 56° (alternate angles)

∠OPQ = 52° and PQ ∥ (dashed line at O)
Angle between the dashed line at O and OP = 52° (alternate angles)

∠NOP = 56° + 52° = 108°

Why it happens: The zig-zag path L–M–N–O–P–Q keeps changing direction, and there is no single transversal joining all the angles. Drawing the parallel helpers gives every corner the same reference direction, so each angle can be split into pieces that transfer from corner to corner.

Tip: This "draw a parallel through the corner" trick is the standard way to handle a zig-zag between two parallel lines. Practise it — it returns in later chapters.

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