How many angles do two intersecting lines form? They form four angles.

Can two straight lines intersect at more than one point? No, two distinct straight lines can intersect at exactly one point.

In Fig 5.2, if ∠a = 120°, find the other angles.

• ∠a and ∠b form a linear pair: ∠b = 180° – 120° = 60°.
• ∠a and ∠c are vertically opposite: ∠c = ∠a = 120°.
• ∠b and ∠d are vertically opposite: ∠d = ∠b = 60°.

Figure it Out (Fig 5.3):

• Linear Pairs: (∠a, ∠b), (∠b, ∠c), (∠c, ∠d), (∠d, ∠a).
• Vertically Opposite Angles: (∠a, ∠c) and (∠b, ∠d).

Can you draw lines where all four angles are equal? Yes. If all four are equal, and they sum to 360°, each angle must be 360°/4 = 90°. These are called perpendicular lines.

Which pairs of lines appear parallel in Fig 5.6? Based on visual inspection, the pairs are (a, b) and (g, h).

Is it possible for all eight angles from a transversal to have different measurements? No. Because of vertically opposite angles (which are equal) and linear pairs (which add to 180°), there can be a maximum of only two distinct angle measures if the lines are not parallel, and only one distinct measure if the lines are perpendicular (all 90°).

Activity 6 (Fig 5.25): If ∠f = 120°, find its alternate angle ∠d.

• ∠f and ∠b are corresponding angles. If the lines were parallel, ∠b = 120°.
• ∠b and ∠d are vertically opposite angles, so ∠d = ∠b.
• Therefore, ∠d = 120°. This shows that if lines are parallel, alternate angles are equal.

Example 1 (Fig 5.26): If lines are parallel and ∠6 = 135°, find other angles.

• Vertically opposite to ∠6 is ∠8, so ∠8 = 135°.
• Corresponding to ∠6 is ∠2, so ∠2 = 135°.
• Vertically opposite to ∠2 is ∠4, so ∠4 = 135°.
• Linear pair with ∠6 is ∠5, so ∠5 = 180° – 135° = 45°.
• All other acute angles (∠1, ∠3, ∠7) will also be 45°.

Example 3 (Fig 5.28): If lines are parallel and ∠3 = 50°, find ∠6.

• Method 1: ∠3 and ∠2 form a linear pair. So ∠2 = 180° – 50° = 130°. ∠2 and ∠6 are corresponding angles, so ∠6 = 130°.
• Method 2: ∠3 and ∠6 are interior angles on the same side of the transversal. Their sum must be 180°. So ∠6 = 180° – 50° = 130°.

a: The 48° angle and ∠a are alternate interior angles. So, a = 48°.

b: The 52° angle and ∠b are interior angles on the same side. So, b = 180° – 52° = 128°.

c: The 99° angle corresponds to the angle vertically opposite to ∠c. So, the angle vertically opposite to ∠c is 99°. Therefore, c = 99°.

d: The angles 99° and 81° form a linear pair (180°), so the lines are straight. The angle vertically opposite to ∠d is 99°. So, d = 99°.

e: The 69° angle and the angle vertically opposite to ∠e form a linear pair with an angle corresponding to 97°. This is complex. Let’s find the angle inside the triangle: 180° – 97° – 69° = 14°. So the angle next to e is 14°. ∠e forms a linear pair with it. e = 180°-14°=166°. No, that is not right. Let’s use a different approach. The angle corresponding to 69° is inside the triangle. The exterior angle is 97°. So, 97° = 69° + ∠e’s vertically opposite angle. So ∠e’s vertically opposite is 28°. e = 28°. This is too advanced. Let’s use simpler rules. The angle adjacent to 97° is 180°-97°=83°. This angle corresponds to the angle inside the triangle on the right. So we have a triangle with angles 69°, 83°, and C. C = 180-69-83 = 28°. ∠e is vertically opposite to C. So e = 28°.

h: This forms a “Z” shape (alternate interior angles). The angle inside the triangle at the top is 75°. The other base angle is vertically opposite to the unknown angle next to 120°. That angle is 180-120=60°. So the triangle has angles 75°, 60°, and C. C = 180-75-60=45°. ∠h is vertically opposite to C. So h = 45°.

Q2 (Find ∠a):

• Top-left: The 100° and the angle adjacent to ‘a’ are corresponding angles. So that angle is 100°. ∠a forms a linear pair with it. a = 180° – 100° = 80°.
• Top-right: The angle corresponding to 62° is vertically opposite to ‘a’. Thus, a = 62°.

Q4 (Fig 5.33):

• ∠GEH and ∠ABC are corresponding angles. ∠GEH = ∠ABC = 45°.
• ∠HEF and ∠IKJ are on a straight line GHIJ, but we don’t know if they are parallel. However, GEF is a straight line. ∠GEH and ∠HEF are a linear pair. So ∠HEF = 180° – 45° = 135°.
• ∠FED is vertically opposite to ∠GEH. So ∠FED = 45°.

Q5 (Fig 5.34):

• Since AB || EF, then AE is a transversal. ∠EAB = 90°. The angle sum of interior angles is 180°. So ∠AEF + ∠EAB = 180°. No. Let’s use the property that if a line is perpendicular to one of two parallel lines, it is perpendicular to the other. Since AE ⊥ AB and AB || CD || EF, then AE ⊥ EF. Thus, ∠AEF = 90°.
• We are given ∠BEF = 55°. We know ∠AEF = 90°. So, y = ∠AEB = ∠AEF – ∠BEF = 90° – 55° = 35°.
• Now consider AB || CD. AE is a transversal. ∠DCE is not helpful. Let’s use transversal BE. ∠ABE and ∠BED are alternate interior angles. So ∠ABE = ∠BED. Not simple. Let’s use the zig-zag rule. The sum of angles pointing one way equals the sum of angles pointing the other. x + y = 55°. We found y=35°. So x + 35° = 55°. x = 20°.