RS Aggarwal Class 7 Mathematics Solutions Chapter 13 Lines and Angles

Question 1. Find the complement of each of the following angles:
(i) 35°
(ii) 47°
(iii) 60°
(iv) 73°
Answer: An angle's complement is obtained by subtracting it from 90°.
(i) Complement of 35° = 90° - 35° = 55°
(ii) Complement of 47° = 90° - 47° = 43°
(iii) Complement of 60° = 90° - 60° = 30°
(iv) Complement of 73° = 90° - 73° = 17°
In simple words: To find a complement, subtract the angle from 90°. The two angles together will always equal 90°.

Exam Tip: Remember that complementary angles always add up to 90°. Keep this relationship clear when solving such problems quickly.

Question 2. Find the supplement of each of the following angles:
(i) 80°
(ii) 54°
(iii) 105°
(iv) 123°
Answer: An angle's supplement is found by subtracting it from 180°.
(i) Supplement of 80° = 180° - 80° = 100°
(ii) Supplement of 54° = 180° - 54° = 126°
(iii) Supplement of 105° = 180° - 105° = 75°
(iv) Supplement of 123° = 180° - 123° = 57°
In simple words: To find a supplement, subtract the angle from 180°. The two angles combined will always give 180°.

Exam Tip: Supplementary angles always sum to 180°. Write this equation first to avoid calculation errors.

Question 3. Two supplementary angles are such that the measure of the larger angle is 36° more than the smaller angle. Find both the angles.
Answer: Let the smaller angle be x. Then the larger angle is x + 36°. Since they are supplementary, their sum equals 180°.

x + (x + 36°) = 180°
2x + 36° = 180°
2x = 144°
x = 72°

Smaller angle = 72°
Larger angle = 72° + 36° = 108°
In simple words: Set up an equation using the given condition. One angle is 36° more than the other, and together they make 180°. Solve to find both angles.

Exam Tip: Always verify your answer by adding both angles to confirm they equal 180°. This quick check prevents errors.

Question 4. An angle is equal to its own supplement. Find the angle.
Answer: Let the angle be x. Since it equals its own supplement:

x = 180° - x
2x = 180°
x = 90°

The required angle is 90°.
In simple words: An angle that matches its supplement must be half of 180°, which is 90°. This is a right angle.

Exam Tip: Remember that only a 90° angle can be its own supplement. This is a special property of right angles.

Question 5. Can two acute angles be supplementary? Can two obtuse angles be supplementary? Can two right angles be supplementary?
Answer:
(i) No. If both angles are acute, meaning each is less than 90°, then their total will always be less than 180°, so they cannot form a supplementary pair.

(ii) No. If both angles are obtuse, meaning each is greater than 90°, then their total will always exceed 180°, so they cannot form a supplementary pair.

(iii) Yes. If both angles are right angles, meaning each measures 90°, then their sum equals 90° + 90° = 180°, so they do form a supplementary pair.
In simple words: Two acute angles add to less than 180°. Two obtuse angles add to more than 180°. Only two right angles add to exactly 180°.

Exam Tip: Always recall the ranges of acute (less than 90°), right (exactly 90°), and obtuse (greater than 90°) angles when testing supplementary relationships.

Question 6. In the given figure, find the value of x.
Answer: Using the linear pair property:

\( \angle AOC + \angle COB = 180° \)
64° + x = 180°
x = 180° - 64°
x = 116°

The value of x is 116°.
In simple words: When two angles form a straight line, they add up to 180°. Subtract the known angle from 180° to get the unknown angle.

Exam Tip: The linear pair property states that adjacent angles on a straight line are supplementary. Use this to quickly solve for unknown angles.

Question 7. Two angles on a straight line are (2x - 10)° and (3x + 20)°. Find both the angles.
Answer: Since the angles lie on a straight line, they form a linear pair and their sum is 180°.

(2x - 10)° + (3x + 20)° = 180°
5x + 10° = 180°
5x = 170°
x = 34°

\( \angle AOC = (2x - 10)° = (2 \times 34 - 10)° = 58° \)
\( \angle BOC = (3x + 20)° = (3 \times 34 + 20)° = 122° \)
In simple words: Add the two angle expressions and set them equal to 180°. Solve for x, then substitute back to find each angle.

Exam Tip: Verify that your angles sum to 180° before finalizing. This prevents careless arithmetic mistakes.

Question 8. In the given figure, a straight line AOB passes through O. Find the value of x.
Answer: Since AOB forms a straight line, all angles around point O on one side must sum to 180°.

\( \angle AOC + \angle BOD + \angle COD = 180° \)
65° + 70° + x° = 180°
135° + x° = 180°
x° = 45°

The value of x is 45°.
In simple words: When a straight line is present, all angles on one side of it add to 180°. Add the known angles and subtract from 180° to find the unknown angle.

Exam Tip: Identify which angles are on the same side of the straight line. Only those angles sum to 180°, not angles on opposite sides.

Question 9. In the given figure, find the value of x and then find all the angles.
Answer: The sum of all angles around a point is 360°.

\( \angle AOB + \angle BOC + \angle COD + \angle DOA = 360° \)
56° + 100° + x° + 74° = 360°
230° + x° = 360°
x° = 130°

Therefore, x = 130°.
In simple words: All angles around a single point total 360°. Add the given angles, subtract from 360°, and you get the missing angle.

Exam Tip: Always use the 360° rule for angles around a point. It is the fastest way to find an unknown angle in such problems.

Question 10. In the given figure, find \( \angle POR \), \( \angle QOR \), and \( \angle QOS \).
Answer:
(i) \( \angle POS \) and \( \angle POR \) form a linear pair, so:
\( \angle POS + \angle POR = 180° \)
114° + \( \angle POR \) = 180°
\( \angle POR \) = 180° - 114° = 66°

(ii) Since \( \angle POS \) and \( \angle QOR \) are vertically opposite angles, they are equal.
\( \angle QOR \) = 114°

(iii) Since \( \angle POR \) and \( \angle QOS \) are vertically opposite angles, they are equal.
\( \angle QOS \) = 66°
In simple words: Use the linear pair property to find one angle. Then use vertically opposite angles, which are always equal, to find the remaining angles.

Exam Tip: Vertically opposite angles are always equal. This property helps solve for unknown angles quickly without extra calculation.

Question 11. In the given figure, find the value of x.
Answer: The sum of all angles around a point is 360°.

\( \angle AOB + \angle BOC + \angle COD + \angle DOA = 360° \)
56° + 100° + x° + 74° = 360°
230° + x° = 360°
x° = 130°

Therefore, x = 130°.
In simple words: Angles formed at a point total 360°. Add all known angles and subtract from 360° to find the unknown angle.

Exam Tip: When multiple angles meet at a point, always apply the 360° rule. Mark each angle clearly to avoid counting any angle twice.