Std 7 Mathematics · Chapter 4

4. Angles and Pairs of Angles

Step-by-step solutions to Practice Sets 15 to 21. Maharashtra State Board (Balbharati), Grade Seven Mathematics.

Practice Set 15

1Observe the figure and complete the table for ∠AWB.

For ∠AWB	Points
Points in the interior	N, R, X, C
Points in the exterior	T, U, Q, V, Y
Points on the arms of the angle	A, G, B (and the vertex W)

Q and V lie just below arm WB, and Y lies beyond it to the right but not on the ray, so all three are in the exterior. Look closely at your textbook figure: a point is on an arm only if the dot sits on the line itself.

2Name the pairs of adjacent angles in the figures below.

First figure (rays NA, NB, NC):

∠ANB and ∠BNC (common arm NB)
∠BNC and ∠CNA (common arm NC)
∠CNA and ∠ANB (common arm NA)

Second figure:

∠PQT and ∠PQR (common vertex Q, common arm QP)

In the second figure, ∠RTS has its vertex at T, not at Q, so it cannot be adjacent to either angle at Q.

3Are the following pairs adjacent angles? If not, state the reason. (i) ∠PMQ and ∠RMQ (ii) ∠RMQ and ∠SMR (iii) ∠RMS and ∠RMT (iv) ∠SMT and ∠RMS

In the figure, the rays from M in order are MP, MT, MQ, MS, MR and MN.

(i) ∠PMQ and ∠RMQ — Adjacent
Common vertex M, common arm MQ; ray MP and ray MR are on opposite sides of MQ, so the interiors are separate.
(ii) ∠RMQ and ∠SMR — Not adjacent
Common arm MR, but ray MS lies inside ∠RMQ. Both MQ and MS are on the same side of MR, so the interiors overlap.
(iii) ∠RMS and ∠RMT — Not adjacent
Common arm MR, but ray MS lies inside ∠RMT, so ∠RMS is a part of ∠RMT. The interiors are not separate.
(iv) ∠SMT and ∠RMS — Adjacent
Common vertex M, common arm MS; ray MT and ray MR are on opposite sides of MS, so the interiors are separate.

Practice Set 16

1The measures of some angles are given below. Write the measures of their complementary angles. (i) 40° (ii) 63° (iii) 45° (iv) 55° (v) 20° (vi) 90° (vii) x°

Complement of an angle = 90° − (measure of the angle).

(i) 90° − 40° = 50°
(ii) 90° − 63° = 27°
(iii) 90° − 45° = 45°
(iv) 90° − 55° = 35°
(v) 90° − 20° = 70°
(vi) 90° − 90° = 0°
(vii) 90° − x° = (90 − x)°

The complement of 45° is 45° itself, and the complement of 90° is 0°.

2(y − 20)° and (y + 30)° are the measures of complementary angles. Find the measure of each angle.

(y − 20) + (y + 30) = 90
2y + 10 = 90
2y = 80
y = 40

∴ y − 20 = 40 − 20 = 20°
∴ y + 30 = 40 + 30 = 70°

Check: 20° + 70° = 90°.

Practice Set 17

1Write the measures of the supplements of the angles given below. (i) 15° (ii) 85° (iii) 120° (iv) 37° (v) 108° (vi) 0° (vii) a°

Supplement of an angle = 180° − (measure of the angle).

(i) 180° − 15° = 165°
(ii) 180° − 85° = 95°
(iii) 180° − 120° = 60°
(iv) 180° − 37° = 143°
(v) 180° − 108° = 72°
(vi) 180° − 0° = 180°
(vii) 180° − a° = (180 − a)°

2The measures of some angles are given below. Use them to make pairs of complementary and supplementary angles. m∠B = 60°, m∠N = 30°, m∠Y = 90°, m∠J = 150°, m∠D = 75°, m∠E = 0°, m∠F = 15°, m∠G = 120°

Pairs of complementary angles (sum 90°):

∠B and ∠N (60° + 30° = 90°)
∠Y and ∠E (90° + 0° = 90°)
∠D and ∠F (75° + 15° = 90°)

Pairs of supplementary angles (sum 180°):

∠B and ∠G (60° + 120° = 180°)
∠N and ∠J (30° + 150° = 180°)

∠D (75°), ∠E (0°) and ∠F (15°) have no supplementary partner in the list, since 105°, 180° and 165° are not given.

3In ΔXYZ, m∠Y = 90°. What kind of a pair do ∠X and ∠Z make?

m∠X + m∠Y + m∠Z = 180° (sum of the angles of a triangle)
m∠X + 90° + m∠Z = 180°
∴ m∠X + m∠Z = 90°

∠X and ∠Z are complementary angles.

4The difference between the measures of the two angles of a complementary pair is 40°. Find the measures of the two angles.

Let the smaller angle be x°. Then the bigger angle is (x + 40)°.

x + x + 40 = 90
2x = 50
x = 25

The angles are 25° and 65°.

Check: 25° + 65° = 90° and 65° − 25° = 40°.

5□PTNM is a rectangle. Write the names of the pairs of supplementary angles.

Each angle of a rectangle is a right angle: m∠P = m∠T = m∠N = m∠M = 90°. The sum of any two of them is 90° + 90° = 180°, so every pair of angles is supplementary.

∠P and ∠T
∠T and ∠N
∠N and ∠M
∠M and ∠P
∠P and ∠N
∠T and ∠M

6*If m∠A = 70°, what is the measure of the supplement of the complement of ∠A?

Complement of ∠A = 90° − 70° = 20°
Supplement of 20° = 180° − 20° = 160°

7If ∠A and ∠B are supplementary angles and m∠B = (x + 20)°, then what would be m∠A?

m∠A + m∠B = 180°
m∠A = 180 − (x + 20)
m∠A = 180 − x − 20
∴ m∠A = (160 − x)°

Practice Set 18

1Name the pairs of opposite rays in the figure alongside.

Lines LM and NT intersect at P. The pairs of opposite rays are:

ray PL and ray PM
ray PN and ray PT

Each pair has the common origin P and together forms a straight line.

2Are the ray PM and PT opposite rays? Give reasons for your answer.

No. Ray PM and ray PT have the same origin P, but they do not form a straight line — PM lies on line LM and PT lies on line NT. They form ∠MPT instead. The ray opposite to PM is PL, and the ray opposite to PT is PN.

Practice Set 19

Draw the pairs of angles as described below. If that is not possible, say why.

(i)Complementary angles that are not adjacent.

Possible. Draw ∠ABC = 30° and ∠PQR = 60° separately. Their sum is 90°, so they are complementary, but they have no common vertex or arm, so they are not adjacent.

30° + 60° = 90°, drawn apart

(ii)Angles in a linear pair which are not supplementary.

Not possible. In a linear pair, the two outer arms form a straight line, so the sum of the two angles is always 180°. Angles in a linear pair are therefore always supplementary.

(iii)Complementary angles that do not form a linear pair.

Possible — in fact every complementary pair is like this. Complementary angles add up to 90°, while angles in a linear pair add up to 180°, so complementary angles can never form a linear pair. Example: adjacent angles of 40° and 50° together make a right angle, not a straight angle.

∠COB = 50°, ∠BOA = 40°; together 90°

(iv)Adjacent angles which are not in a linear pair.

Possible. Draw two adjacent angles whose sum is not 180°, for example ∠QOP = 30° and ∠ROQ = 50°. They share vertex O and arm OQ, but rays OP and OR do not form a straight line.

Adjacent angles 30° and 50° (sum 80°, not a linear pair)

(v)Angles which are neither complementary nor adjacent.

Possible. Draw ∠ABC = 70° and ∠PQR = 50° separately. Their sum is 120°, not 90°, and they have no common vertex or arm.

70° + 50° = 120°, drawn apart

(vi)Angles in a linear pair which are complementary.

Not possible. Angles in a linear pair always add up to 180°, but complementary angles must add up to 90°. Both cannot be true at once.

Practice Set 20

1Lines AC and BD intersect at point P. m∠APD = 47°. Find the measures of ∠APB, ∠BPC, ∠CPD.

∠APD and ∠APB are angles in a linear pair.
m∠APB = 180° − 47° = 133°

∠BPC is vertically opposite to ∠APD.
∴ m∠BPC = 47°

∠CPD is vertically opposite to ∠APB.
∴ m∠CPD = 133°

Check: 47° + 133° + 47° + 133° = 360°.

2Lines PQ and RS intersect at point M. m∠PMR = x°. What are the measures of ∠PMS, ∠SMQ and ∠QMR?

∠PMR and ∠PMS are angles in a linear pair.
m∠PMS = (180 − x)°

∠SMQ is vertically opposite to ∠PMR.
∴ m∠SMQ = x°

∠QMR is vertically opposite to ∠PMS.
∴ m∠QMR = (180 − x)°

Practice Set 21

1∠ACD is an exterior angle of ΔABC. The measures of ∠A and ∠B are equal. If m∠ACD = 140°, find the measures of the angles ∠A and ∠B.

An exterior angle equals the sum of its remote interior angles.

m∠A + m∠B = m∠ACD = 140°
Since m∠A = m∠B, 2 × m∠A = 140°
∴ m∠A = 70°

m∠A = 70° and m∠B = 70°.

2Using the measures of the angles given in the figure alongside, find the measures of the remaining three angles.

Ray OF and ray OC form a straight line, so the three angles above line FC add up to 180°.

4y + 8y + 6y = 180
18y = 180
y = 10

m∠FOA = 4y = 40°, m∠AOB = 8y = 80°, m∠BOC = 6y = 60°

The remaining three angles are vertically opposite to these:

m∠COD = m∠FOA = 40°
m∠DOE = m∠AOB = 80°
m∠EOF = m∠BOC = 60°

Lines AD, BE and FC meet at O. With y = 10: ∠FOA = 4y = 40°, ∠AOB = 8y = 80°, ∠BOC = 6y = 60°; the angles below FC are vertically opposite to these.

3*In the isosceles triangle ABC, ∠A and ∠B are equal. ∠ACD is an exterior angle of ΔABC. The measures of ∠ACB and ∠ACD are (3x − 17)° and (8x + 10)° respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B.

∠ACB and ∠ACD are angles in a linear pair.

(3x − 17) + (8x + 10) = 180
11x − 7 = 180
11x = 187
x = 17

m∠ACB = 3 × 17 − 17 = 34°
m∠ACD = 8 × 17 + 10 = 146°

m∠A + m∠B = m∠ACD = 146° (exterior angle property)
Since m∠A = m∠B, each = 146° ÷ 2 = 73°

m∠A = 73° and m∠B = 73°.

Check: 73° + 73° + 34° = 180°.

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