Class 9 Mathematics · 37 questions · 18 with answers
If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, then the greater of the two angles is
Solution : Answer (B)
Let OA, OB, OC and OD are rays in the anticlockwise direction such that ∠ AOB = ∠COD = 100°, ∠BOC = 82° and ∠AOD = 78°. Is it true to say that AOC and BOD are lines?
Solution : AOC is not a line, because ∠ AOB + ∠ COB = 100° + 82° = 182°, which is not equal to 180°. Similarly, BOD is also not a line.
A transversal intersects two lines in such a way that the two interior angles on the same side of the transversal are equal. Will the two lines always be parallel? Give reason for your answer.
Solution : In general, the two lines will not be parallel, because the sum of the two equal angles will not always be 180°. Lines will be parallel when each equal angle is equal to 90°.
In Fig. 6.7, AB, CD and EF are three lines concurrent at O. Find the value of y.
Solution : ∠AOE = ∠BOF = 5y (Vertically opposite angles) Also, ∠COE + ∠AOE + ∠AOD = 180° So, 2y + 5y + 2y = 180° or, 9y = 180°, which gives y = 20°. Fig. 6.7
In Fig.6.8, x = y and a = b. Prove that l || n.
Solution: x = y (Given) Therefore, l || m (Corresponding angles) (1) Also, a = b (Given) Therefore, n || m (Corresponding angles) (2) From (1) and (2), l || n (Lines parallel to the same line) Fig. 6.8
In Fig. 6.15, m and n are two plane mirrors perpendicular to each other. Show that incident ray CA is parallel to reflected ray BD. Fig. 6.15
Solution: Let normals at A and B meet at P. As mirrors are perpendicular to each other, therefore, BP || OA and AP || OB. So, BP ⊥ PA, i.e., ∠ BPA = 90° Therefore, ∠ 3 + ∠ 2 = 90° (Angle sum property) (1) Also, ∠1 = ∠2 and ∠4 = ∠3 (Angle of incidence = Angle of reflection) Therefore, ∠1 + ∠4 = 90° [From (1)] (2) Adding (1) and (2), we have ∠1 + ∠2 + ∠3 + ∠4 = 180° i.e., ∠CAB + ∠DBA = 180° Hence, CA || BD LINES AND ANGLES 61
Prove that the sum of the three angles of a triangle is 180°.
Solution: See proof of Theorem 6.7 in Class IX Mathematics Textbook.
Bisectors of angles B and C of a triangle ABC intersect each other at the point O. Prove that ∠BOC = 90° + ∠A.
Solution: Let us draw the figure as shown in Fig. 6.16 ∠A + ∠ABC + ∠ACB = 180° (Angle sum property of a triangle) Fig. 6.16
In Fig. 6.1, if AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then ∠QRS is equal
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(C) 145°
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
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(D) a right triangle
An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is 1° 1° 1°
(A) 37
The angles of a triangle are in the ratio 5 : 3 : 7. The triangle is
(A) an acute angled triangle
If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be
(D) 155°
In Fig. 6.2, POQ is a line. The value of x is
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(A) 20°
In Fig. 6.3, if OP||RS, ∠OPQ = 110° and ∠QRS = 130°, then ∠ PQR is equal to
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(C) 60°
Angles of a triangle are in the ratio 2 : 4 : 3. The smallest angle of the triangle is
(B) 40°
For what value of x + y in Fig. 6.4 will ABC be a line? Justify your answer.
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x + y must be equal to 180°. For ABC to be a line, the sum of the two adjacent angles must be 180°.
Can a triangle have all angles less than 60°? Give reason for your answer.
No, angle sum will be less than 180°.
Can a triangle have two obtuse angles? Give reason for your answer.
No, angle sum cannot be more than 180°.
How many triangles can be drawn having Fig. 6.4 its angles as 45°, 64° and 72°? Give reason for your answer. LINES AND ANGLES 57
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None, angle sum cannot be 181°.
How many triangles can be drawn having its angles as 53°, 64° and 63°? Give reason for your answer.
Infinitely many triangles. sum of the angles of every triangle is 180°.
In Fig. 6.5, find the value of x for which the lines l and m are parallel.
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136°.
Two adjacent angles are equal. Is it necessary that each of these angles will be a right angle? Justify your answer. Fig. 6.5
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No, each of these will be a right angle only when they form a linear pair.
If one of the angles formed by two intersecting lines is a right angle, what can you say about the other three angles? Give reason for your answer.
Each will be a right angle. Linear pair axiom .
In Fig.6.6, which of the two lines are parallel and why? Fig. 6.6
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l || m because 132° + 48° = 180°, p is not parallel to q, because 73° + 106° ≠ 180°.
Two lines l and m are perpendicular to the same line n. Are l and m perpendicular to each other? Give reason for your answer.
No, they are parallel
In Fig. 6.9, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and OD ⊥ OE. Show that the points A, O and B are collinear. Fig. 6.9
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In Fig. 6.10, ∠1 = 60° and ∠6 = 120°. Show that the lines m and n are parallel. Fig. 6.10
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AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (Fig. 6.11). Show that AP || BQ. Fig. 6.11 LINES AND ANGLES 59
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If in Fig. 6.11, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m.
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In Fig. 6.12, BA || ED and BC || EF. Show that ∠ABC = ∠DEF [Hint: Produce DE to intersect BC at P (say)]. Fig. 6.12
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In Fig. 6.13, BA || ED and BC || EF. Show that ∠ ABC + ∠ DEF = 180° Fig. 6.13
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In Fig. 6.14, DE || QR and AP and BP are bisectors of ∠ EAB and ∠ RBA, respectively. Find ∠APB. Fig. 6.14
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The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles of the triangle.
A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that ∠ BAL = ∠ ACB.
Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.
1 1 1 Therefore, ∠A + ∠ABC + ∠ACB = × 180° = 90°
2 2 2 i.e., ∠A + ∠OBC + ∠OCB = 90° (Since BO and CO are bisectors of ∠B and ∠C) (1) But ∠BOC + ∠OBC + ∠OCB =180° (Angle sum property) (2) Subtracting (1) from (2), we have ∠BOC + ∠OBC + ∠OCB – ∠A – ∠OBC – ∠OCB = 180° – 90° i.e., ∠BOC = 90° + ∠A
If two lines intersect, prove that the vertically opposite angles are equal.
Bisectors of interior ∠B and exterior ∠ACD of a ∆ ABC intersect at the point T. Prove that ∠ BTC = ∠ BAC.
A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.
Prove that through a given point, we can draw only one perpendicular to a given line. [Hint: Use proof by contradiction].
Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other. [Hint: Use proof by contradiction].
Prove that a triangle must have atleast two acute angles.
In Fig. 6.17, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that ∠APM = ( ∠Q – ∠R). Fig. 6.17
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