Haryana State Board HBSE 9th Class Maths Important Questions Chapter 6 Lines and Angles Important Questions and Answers.

## Haryana Board 9th Class Maths Important Questions Chapter 6 Lines and Angles

Very Short Answer Type Questions

Question 1.

Find the complement of the following angles :

(i) 64° 23′ 35″

(ii) 28° 25′ 48″.

Solution :

We can write,

90° = 89° 59′ 60″

(i) Complement of 64° 23′ 35″ = An angle of (90° – 64° 23′ 35″) = An angle of 25° 36′ 25″.

(ii) Complement of 28° 25′ 48″ = An angle of (90°- 28 °25′ 48″) = An angle of 61° 34′ 12″.

Question 2.

Find the measures of suppliment of the following angles :

(i) 115° 35′ 42″

(ii) 85° 15′ 59″.

Solution :

We can write,

180° = 179° 59′ 60″

(i) Supplement of 115° 35′ 42″ = An angle of (180° – 115°35′ 42″) = An angle of 64° 24′ 18″.

(ii) Suppliment of 85° 15′ 59″ = An angle of (180° – 85° 15′ 59”) = An angle of 94° 44′ 01″.

Question 3.

Find the measure of an angle which is 34° more than its complement.

Solution :

Let the complement of an angle be x° then

measure of an angle = (x + 34)°

We know that,

sum of an angle and its complement = 90°

∴ x° + (x + 34)° = 90°

⇒ 2x° + 34° = 90°

⇒ 2x° = 90° – 34°

⇒ 2x° = 56°

⇒ x° = \(\frac {56°}{2}\) = 28°

Hence,measure of an angle

= 28° + 34° = 62°.

Question 4.

Find the measure of an angle which is 26° less than its supplement.

Solution :

Let the supplement of an angle be x°

Then measure of an angle = (x – 26)°

We know that,

Sum of an angle and its supplement = 180°

∴ x° + (x – 26)° = 180°

⇒ 2x° – 26° = 180°

⇒ 2x° = 180° + 26°

⇒ 2x° = 206°

⇒ x° = \(\frac {206°}{2}\) = 103°

Hence measure of an angle = 103° – 26° = 77°.

Question 5.

Find the angle, which is 18° less than five times its complement.

Solution :

Let the complement angle be x°

Five times of complement angle = 5x°

Then angle = (5x – 18)°

We know that, Sum of an angle and its complement = 90°

∴ (5x – 18)° + x° = 90°

⇒ 6x° – 18° = 90°

⇒ 6x° = 90° + 18°

⇒ 6x° = 108°

⇒ x° = \(\frac {108°}{6}\) = 18°

Hence,measure of angle

= 5 × 18° – 18°

= 90° – 18° = 72°.

Question 6.

Find the angle, which is four times its supplement

Solution :

Let the supplement angle be x°.

Then angle = 4x°

We know that, Sum of an angle and its suppliment = 180°

∴ 4x° + x° = 180°

⇒ 5x° = 180°

⇒ x° = \(\frac {180°}{5}\) = 18°

Hence,measure of the angle

= 4 × 36° = 144°.

Question 7.

Find the measure of an angle whose supplement is three times of its complement.

Solution :

Let the angle be x°, then Supplement of an angle = (180 – x)°

Compliment of an angle = (90 – x)°

According to question,

(180 – x)° = 3(90 – x)°

⇒ 180° – x° = 270° – 3x°

⇒ – x° + 8x° = 270° – 180°

⇒ 2x° = 90°

⇒ x° = \(\frac {90°}{2}\) = 45°

Hence, the measure of the angle = 45°

Question 8.

Find the measure of an angle, if five times its complement is equal to two times of its supplement.

Solution :

Let the angle be x°, then

Complement of an angle = (90 – x)°

Supplement of an angle = (180 – x)°

According to question,

5(90 – x)° = 2(180 – x)°

⇒ 450° – 5x° = 360° – 2x°

⇒ – 5x° + 2x° = 360° – 450°.

⇒ – 3x° = – 90°

⇒ x° = \(\frac {-90°}{- 3}\) = 30°

Hence the measure of the angle = 30°.

Short Answer Type Questions

Question 1.

One of two complementary angles is seven-eighth as large as the other. How many degree are in each angle?

Solution :

Let the other complementary angle be x°.

One complimentary angle = \(\frac {7}{8}\)x°

We know that, Sum of two complementary angles = 90°

Hence, other complementary angle = 48° and one complementary angle = \(\frac {7}{8}\) × 48° = 42°.

Question 2.

In the figure, AC and BC are opposite rays. If 2x – 3y = 60°, then find the values of x and y.

Solution :

Since, AC and BC are opposite rays. Therefore, ∠ACD + ∠BCD = 180°,

(Linear pair axiom)

⇒ x + y = 180° …(i)

2x – 3y = 60° (Given) …(ii)

Multiplying equation (i) by 3 and adding equation (i) and (ii), we get

⇒ x = \(\frac {600°}{5}\) = 120°

Substituting the value of x in the equation (i), we get

120° + y = 180°

⇒ y = 180° – 120° = 60°

Hence, x = 120° and y = 60°

Question 3.

In figure, BCA is a line. Find the value of x.

Solution :

Since, BCA is a line. Therefore ∠BCD + ∠ACD = 180°,

(Linear pair axiom)

⇒ ∠BCD + ∠DCE + ∠ACE = 180°,

[∵ ∠ACD = ∠DCE + ∠ACE]

⇒ 2x° + (2x° + 30)° + x° = 180°

⇒ 5x° + 30° = 180°

⇒ 5x° = 180° – 30°

⇒ 5x° = 150°

⇒ x° = \(\frac {150°}{5}\) = 30°

Hence, x° = 30°

Question 4.

In figure, PQR is a line. If a : b : c = 2 : 3 : 5, find the values of a, b and c.

Solution :

Since, PQR is a line. Therefore ∠PQM + ∠RQM = 180°, (Linear pair axiom)

⇒ ∠PQM + ∠MQS + ∠RQS = 180°,

[∵ ∠RQM = ∠MQS + ∠RQS]

⇒ a + c + b = 180°

But a : b : c = 2 : 3 : 5

Sum of ratios = 2 + 3 + 5 = 10

∴ a = \(\frac {2}{10}\) × 180° = 36°

b = \(\frac {3}{10}\) × 180° = 54°

and c = \(\frac {5}{10}\) × 180° = 90°

Hence, a = 36°, b = 54 and c = 90°.

Question 5.

In figure, three straight lines AB, CD and EF intersect at a point o forming the angles as shown. Find the values of x, y, z and r.

Solution:

y = 60°, (Vertically opposite angles)

r = 90° (Vertically opposite angles)

Since, AOB is a line.

∠AOE + ∠BOE = 180°

(Linear pair axiom)

⇒ ∠AOD + ∠DOE + ∠BOE = 180°

⇒ y + x + 90° = 180°

⇒ 60° + x + 90° = 180°

⇒ x + 150° = 180°

⇒ x = 180° – 150°= 30°

z = x

(Vertically opposite angles)

⇒ z = 30°

Hence,

x = 30°, y = 60°,

z = 30°, r = 90°.

Question 6.

In the figure , AB || CD, PQ || AX and ∠BAC = 70°, find ∠DPQ

Solution :

∵ AB || CD and AX is a transversal

∴ ∠BAC + ∠ACD = 180°,

(Sum of a pair of allied angles is 180°)

⇒ 70° + ∠ACD = 180°,(∵ ∠BAC = 70°)

⇒ ∠ACD = 180° – 70°

⇒ ∠ACD = 110°

Now PQ || AX and CD is a transversal.

∠DPQ = ∠ACD, (Corresponding angles axiom)

⇒ ∠DPQ = 110°

Hence, ∠DPQ = 110°

Question 7.

In the figure, AB || CD aud m || n, find the value of x.

Solution :

∠QPR = 3x,

(Vertically opposite angles)

∵ AB || CD and line m is a transversal.

∴ ∠QPR + ∠PRS = 180°,

(A pair of allied angles is supplementary)

⇒ 3x + ∠PRS = 180°

⇒ ∠PRS = 180° – 3x ……(i)

Now, m || n and line CD is a transversal.

∠PRS = ∠QSD,

(Corresponding angles axiom)

⇒ 180° – 3x = 2x + 5°, [From (i), ∠PRS = 180° – 3x and ∠QSD = 2x + 5°]

⇒ 180° – 5° = 2x + 3x

⇒ 175° = 5x

⇒ \(\frac {175°}{5}\) = x

⇒ x = 35°

Hence, x = 35°

Question 8.

In the figure, AB || CD, find the value of x.

Solution :

From E, draw EF || AB || CD.

Now, AB || EF and AE is the transversal.

∴ ∠BAE + ∠AEF = 180°

(A pair of allied angles are supplementary)

⇒ 110° + ∠AEC + ∠CEF = 180°

⇒ 110° + 25° + ∠CEF = 180°

⇒ 135° + ∠CEF = 180°

⇒ ∠CEF = 180° – 135°

⇒ ∠CEF = 45°

Again CD || EF and CE is the transversal.

∴ ∠DCE + ∠CEF = 180°,

(A pair of allied angles is supplementary)

⇒ x + 45° = 180°

⇒ x = 180° – 45°

⇒ x = 135°

Hence, x = 135°

Question 9.

In the figure, AB || DE and x = y, prove that BC || EF.

Solution :

Since AB || DE and BC is the transversal.

Therefore, ∠DOC = ∠ABC,

(Corresponding angles axiom)

⇒ ∠DOC = x

But x = y, (Given)

∴ ∠DOC = y

Thus, a pair of corresponding angles ∠DOC and y are equal. By converse of corresponding axiom, we have BC || EF. Hence proved

Question 10.

In the given figure, AB || CD, PQ || BC, ∠BAC = 52° and ∠DRS = 43°. Find the value of x.

Solution :

Since, AB || CD and AC is the transversal.

∴ ∠ACD = ∠BAC,

(Alternate interior angles)

⇒ ∠ACD = 52°

⇒ ∠QCR = 52°,

(∵ ∠ACD = ∠QCR) …(i)

∠QRC = ∠DRS,

(Vertically opposite angles)

⇒ ∠QRC = 43° …….(ii)

In ΔCQR, we have

∠AQR = ∠QCR + ∠QRC,

⇒ x = 52° + 43°, [From (i) and (ii), ∠QCR = 52° and ∠QRC = 43°)

⇒ x = 95°

Hence, x = 95°

Question 11.

The sides BC, CA and AB of a triangle ABC are produced in order, forming exterior angles ∠ACD, ∠BAE and ∠CBF. Show that ∠ACD + ∠BAE + ∠CBF = 360°

Solution :

In ΔABC, we have

∠ACD = ∠1 + ∠3 ………….(i)

∠BAE = ∠1 + ∠2 ………….(ii)

∠CBF = ∠2 + ∠3 ………….(iii)

Adding (i), (ii) and (iii), we get

∠ACD + ∠BAE + ∠CBF = ∠1 + ∠3 + ∠1 + ∠2 + ∠2 + ∠3

⇒ ∠ACD + ∠BAE + ∠CBF = 2∠1 + 2∠2 + 2∠3

⇒ ∠ACD + ∠BAE + ∠CBF = 2(∠1 + ∠2 + ∠3)

But ∠1 + ∠2 + ∠3 = 180°,

(Sum of the angles of a triangle = 180°)

∴ ∠ACD + ∠BAE + ∠CBF = 2 × 180°

⇒ ∠ACD + ∠BAE + ∠CBF = 360°

Hence proved

Question 12.

In the given figure, show that ∠A + ∠B + ∠C + ∠P + ∠Q + ∠R = 360°.

Solution :

In ΔABC, we have

∠A + ∠B + ∠C= 180°,

(Sum of angles of a triangle = 180°) …(i)

In ΔPQR, we have

∠P + ∠Q + ∠R = 180°,

(Sum of angles of a triangle = 180°) …(ii)

Adding (i) and (ii), we get

∠A + ∠B + ∠C + ∠P + ∠Q + ∠R = 180° + 180°

⇒ ∠A + ∠B + ∠C + ∠P + ∠Q + ∠R = 360°

Hence proved

Question 13.

ABC is a triangle in which ∠A = 66°, the internal bisectors of ∠B and ∠C intersect at O. Find the measure of ∠BOC.

Solution :

In a triangle ΔABC, we have

∠A + ∠B + ∠C = 180°,

(Sum of angles of a triangle = 180°)

⇒ 66° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° – 66° = 114°

⇒ \(\frac {1}{2}\)∠B + \(\frac {1}{2}\)∠C = \(\frac {114°}{2}\) = 57°

⇒ ∠1 + ∠2 = 57° …….(i)

[∵ BO and CO are bisectors of ∠B and ∠C respectively]

Now, in ΔBOC, we have

∠1 + ∠2 + ∠BOC = 180°,

(Sum of angles of a triangle = 180°)

⇒ 57° + ∠BOC = 180°, [using (i)]

⇒ ∠BOC = 180° – 57°

⇒ ∠BOC = 123°

Hence, ∠BOC = 123°

Question 14.

The sum of two angles of a triangle is equal to its third angle. Prove that triangle is right-angled.

Solution :

Let the smaller angles be A and C and greater angle be B. According to question,

∠B = ∠A + ∠C ……..(i)

But, ∠A + ∠B + ∠C = 180°,

(∵ Sum of angles of a triangle = 180°)

(∠A + ∠C) + ∠B = 180°

⇒ ∠B + ∠B = 180°, [Using (i)]

⇒ 2∠B = 180°

⇒ ∠B = \(\frac {180°}{2}\)

⇒ ∠B = 90°

Hence, triangle ABC is right angled.

Proved

Question 15.

In the given figure, ∠P = ∠Q and ∠1 = ∠2. Prove that RT || PQ.

Solution :

In ΔRPQ, we have

∠SRQ = ∠P + ∠Q

⇒ ∠1 + ∠2 = ∠Q + ∠Q

(∵ It is given ∠P = ∠Q)

⇒ ∠2 + ∠2 = 2∠Q

(∵ It is given ∠1 = ∠2)

⇒ 2∠2 = 2∠Q

⇒ ∠2 = ∠Q

Thus, a pair of alternate interior angles 2 and Q are equal, then

RT || PQ. Hence proved

Question 16.

In figure, p || t and m || n. If ∠1 = 75°, prove that ∠2 = ∠1 + \(\frac {1}{3}\) of a right angle.

Solution :

∵ m || n and line p is a transversal

∴ ∠3 = ∠1

(Corresponding angles)

⇒ ∠3 = 75° (∴ ∠1 = 75°)

Now p || t and line n is a transversal.

∴ ∠2 + ∠3 = 180° (A pair of allied angles is supplementary)

⇒ ∠2 + 75° = 180°

⇒ ∠2 = 180° – 75°

⇒ ∠2 = 105°

⇒ ∠2 = 75° + 30°

⇒ ∠2 = ∠1 + \(\frac {1}{3}\) × 90°

(∵ ∠1 = 75°)

⇒ ∠2 = ∠1 + \(\frac {1}{3}\) of a right angle.

Hence proved

Long Answer Type Questions

Question 1.

In the figure, OP and OQ are respectively bisectors of angles ∠BOD and ∠AOC. Show that the rays OP and OQ are in the same straight line.

Solution :

∠1 + ∠2 + ∠6 + ∠4 + ∠3 + ∠5 = 360° …….(i)

(Sum of angles at a point = 360°)

∵ ray OP is the bisertor of ∠BOD.

∴ ∠3 = ∠4 …….(ii)

∵ ray OQ is the bisector of ∠AOC.

∴ ∠1 = ∠2 …….(iii)

∠5 = ∠6, (Vertically opposite angles) …(iv)

From (i), (ii), (iii) and (iv), we get

∠2 + ∠2 + ∠6 + ∠4 + ∠4 + ∠6 = 360°

⇒ 2∠2 + 2∠6 + 2∠4 = 360°

⇒ 2(∠2 + ∠6 + ∠4) = 360°

⇒ ∠2 + ∠6 + ∠4 = \(\frac {360°}{2}\) = 180°

⇒ ∠POQ = 180°

Hence, OP and OQ are in the same straight line.

Hence Proved

Question 2.

In the figure, AB || CD, find the value of x.

Solution :

Draw a line EF parallel to AB.

∵ AB || EF and a transversal PQ cuts them at P and Q respectively.

∴ ∠APQ + ∠EQP = 180°,

(∵ Sum of a pair of allied angles is 180°)

⇒ 40° + ∠EQP = 180°, (∵ ∠APQ = 40°)

⇒ ∠EQP = 180° – 40°

⇒ ∠EQP = 140° …….(i)

∴ AB || EF, (By construction)

and AB || CD, (Given)

∴ EF || CD, (By theorem 6.6)

∠EQR = ∠QRD,

(Alternate interior angles)

⇒ ∠EQR = 120°,

(∵ ∠QRD = 120°) …(ii)

Adding (i) and (ii), we get

∠EQP + ∠EQR = 140° + 120°

⇒ Reflex angle PQR = 260°

⇒ x = 260°

Hence, x= 260°.

Question 3.

In the given figure :

(i) If x = 70°, find y and z.

(ii) If 3x = 2y, find x.

(iii) If z = 2(x + 15), find y.

Solution :

(i) ∵ CD || EF and CE is a transversal.

∴ x + z = 180° (Sum of a pair of allied angles is 180°)

⇒ 70° + z = 180°, (∵ x = 70°)

⇒ z = 180° – 70°

⇒ z = 110°

∵ AB || CD and AC is a transversal.

∴ y = z,

(Corresponding angles axiom)

⇒ y = 110°

Hence y = 110° and z = 110°.

(ii) 3x = 2y, (Given)

⇒ \(\frac {3}{2}\)x = y …….(i)

∵ AB || EF and AE is a transversal.

∴ x + y = 180° (Sum of a pair of allied angles is 180°)

⇒ x + \(\frac {3}{2}\)x = 180°

[From (i), y = \(\frac {3}{2}\)x]

⇒ \(\frac{2 x+3 x}{2}\) = 180°

⇒ 5x = 2 × 180°

⇒ x = \(\frac {360°}{5}\)

⇒ x = 72°

Hence, x = 72°.

(iii) ∵ CD || EF and CE is a transversal.

∴ x + z = 180°, (Sum of a pair of allied angle is 180°)

⇒ x + 2(z + 15) = 180,[∵ z = 2(x + 15°)

⇒ x + 2x + 30° = 180°

⇒ 3x + 30° = 180°

⇒ 3x = 180° – 30° = 150°

⇒ x = \(\frac {150°}{3}\) = 50°

Now,

z = 2(x + 15°)

⇒ z = 2(50° + 15°)

[Put x = 50°]

⇒ z = 2 × 65°

⇒ z = 130°

Now, y = z,

(Corresponding angles)

⇒ y = 130°

Hence, y = 130°

Question 4.

In the figure, PQ || RS, find the value of x.

Solution :

Through T draw LTM || PQ || RS.

∵ LT || PQ and PT is the transversal.

∴ ∠LTP = ∠TPQ,

(Alternate interior angles)

⇒ ∠LTP = 55° ……..(i)

and TM || RS and TR is the transversal.

∴ ∠MTR + ∠TRS = 180°,

(A pair of allied angles is supplementary)

⇒ ∠MTR + 115° = 180°

⇒ ∠MTR = 180° – 115°

⇒ ∠MTR = 65° ……..(ii)

Now LTM is a line.

∴ ∠LTP + ∠PTR + ∠MTR = 180°

(Straight line)

⇒ 55° + x + 65° = 180°. [From (i) and (ii), ∠LTP = 55° and ∠MTR = 65°]

⇒ 120° + x = 180°

⇒ x = 180° – 120°

⇒ x = 60°

Hence, x = 60°.

Question 5.

In the figure, AB || CD, find the values of x, y and z.

Solution :

∵ AB || CD and EF is the transversal

∴ ∠PQB = ∠EPD,

(Corresponding angles)

⇒ x + 25° = 50°

⇒ x = 50° – 25° = 25° and ∠DPQ + ∠PQB = 180°,

(A pair of allied angles is supplementary)

⇒ (y + 60°) + (x + 25°) = 180°

⇒ y + 60° + 25° + 25° = 180° (∵ x = 25°)

⇒ y + 110°= 180°

⇒ y = 180° – 110°

⇒ y = 70°

In the triangle ΔPQR, Sum of the angles of a triangle is 180°

∴ x + y + z = 180°

⇒ 25° + 70° + z = 180°

⇒ 95° + z = 180°

⇒ z = 180° – 95°

⇒ z = 85°

Hence,

x = 25°, y = 70° and z = 85°

Question 6.

In the given figure, if ΔABC right angled at B, then show that ∠x = ∠y.

Solution :

Since, CDB is a straight line.

∠ADC + ∠ADB= 180°,(Linear pair axiom)

⇒ 110° + ∠ADB = 180°

⇒ ∠ADB = 180° – 110°

⇒ ∠ADB = 70°

In ΔABD, we have

∠ADC = ∠DAB + ∠ABD,

⇒ 110° = ∠y + 90°

⇒ 110° – 90° = ∠y

⇒ ∠y = 20° ………..(i)

Again in ΔADC, we have

∠ADB = ∠ACD + ∠DAC,

(By theorem 6.8)

⇒ 70° = 50° + ∠x

⇒ 70° – 50° = ∠x

⇒ ∠x = 20° ………..(ii)

From (i) and (ii), we get

∠x = ∠y. Hence proved

Question 7.

In the figure 6.110, AB || CD and EF is a transversal. Prove that bisectors of interior angles on the same side of transversal EF intersect at right angle. [NCERT Exemplar Problems]

Solution :

Since AB || CD and EF is the transversal.

∠DPR + ∠PRB = 180°, (A pair of interior angles of the same side of transversal is supplementary) …(i)

⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°

But

∠1 = ∠2, (∵ PQ is the bisector of ∠DPR) …(ii) and

∠3 = ∠4, (∵ RQ is the bisector of ∠PRB) …(iii)

From (i), (ii) and (iii), we get

∠1 + ∠1 + ∠3 + ∠3 = 180°

⇒ 2∠1 + 2∠3 = 180°

⇒ 2(∠1 + ∠3) = 180°

⇒ ∠1 + ∠3 = 90° ……..(iv)

In the ΔPQR, Sum of the angles of a triangle is 180°,

∴ ∠1 + ∠3 + ∠PQR = 180°

⇒ 90° + ∠PQR = 180°,

[From (iv), ∠1 + ∠3 = 90°]

⇒ ∠PQR = 180° – 90°

⇒ ∠PQR = 90°

⇒ ∠PQR = a right angle

Hence, bisectors of interior angles on the same side of the transversal EF intersect at right angle.

Hence Proved

Question 8.

In the given figure, AD || BC, if y^{0} = \(\frac {3}{4}\)x^{0} and x^{0} = \(\frac {4}{5}\)z^{0}, find the values of x^{0}, y^{0}, z^{0} and k^{0}.

Solution :

Since, AD || BC and BD is the transversal.

∴ ∠DBC = ∠ADB,

(Alternate interior angles)

⇒ ∠DBC = x^{0}

In ΔBDC, we have

x° + y° + z° = 180°, (Sum of angles of a triangle = 180°)

Now in ΔDAB,

∠DAB + ∠ADB + ∠ABD = 180°,

(Sum of angles of a triangle = 180°)

⇒ k° + x° + 65° = 180°

⇒ k° + 60° + 65° = 180° (∵ x = 60°)

⇒ k° + 125° = 180°

⇒ k° = 180° – 125°

⇒ k° = 55°

Hence,

x° = 60°, y° = 45°, z° = 75° and k° = 55°

Question 9.

The side BC of a tr iangle ΔABC is produced to D. The bisector of ∠A meets BC at M (see in figure). Prove that ∠ABC + ∠ACD = 2∠AMC.

Solution :

In ΔABC, we have

∠ACD = ∠ABC + ∠BAC,

(By theorem 6.8) …(i)

But AM is the bisector of ∠BAC. (Given)

∴ ∠BAM = ∠CAM

or ∠BAC = 2∠BAM

Putting the value of ∠BAC in (i), we get

∠ACD = ∠ABC + 2∠BAM … (ii)

Again, in ΔABM, we have

∠AMC = ∠ABC + ∠BAM

⇒ 2∠AMC = 2∠ABC + 2∠BAM ……(iii)

Subtracting (ii) from (iii), we get

2∠AMC – ∠ACD = ∠ABC

⇒ 2∠AMC = ∠ABC + ∠ACD

⇒ ∠ABC + ∠ACD = 2∠AMC. Hence proved

Question 10.

If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles enclose a rectangle.

Solution :

Given : Two parallel lines AB and CD are intersected by a transversal EF at P and R respectively. PQ, RS, PS and RQ are the bisectors of the two pairs of interior angles.

To prove: PQRS is a rectangle.

Proof : ∠BPR = ∠PRC,

(A pair of alternate interior angles)

⇒ \(\frac {1}{2}\)∠BPR = \(\frac {1}{2}\)∠PRC

⇒ ∠QPR = ∠PRS

[∵ PQ and RS are bisectors of ∠BPR and ∠PRC

∴ ∠QPR = \(\frac {1}{2}\)∠BPR and ∠PRS = \(\frac {1}{2}\)∠PRC]

But these are alternate interior angles.

PQ || SR, (By theorem 6.3)

Similarly, SP || RQ

∴ PQRS is a parallelogram.

Now, ∠BPR + ∠PRD = 180°,

(Sum of a pair of allied angles = 180°)

⇒ \(\frac {1}{2}\)∠BPR + \(\frac {1}{2}\)∠PRD = \(\frac {180°}{2}\)

⇒ \(\frac {1}{2}\)∠BPR + \(\frac {1}{2}\)∠PRD = 90°

⇒ ∠QPR + ∠QRP = 90° ,

[∵ PQ and RQ are the bisectors of ∠BPR and ∠PRD respectively]

Now in ΔPQR,

∠QPR + ∠PQR + ∠QRP = 180°,

(Sum of angles of a triangle is 180°)

⇒ ∠PQR + 90° = 180°, [Using (i)]

⇒ ∠PQR = 180° – 90° = 90°

In a parallelogram PQRS, ∠PQR = 90°

Hence, PQRS is a rectangle. Hence Proved

Question 11.

In the given figure, AB = AC in ΔABC. From B, BP is drawn such that BP = BC. From P, a line PQ is drawn parallel to BC to meet AB at Q. If exterior ∠RAB = 124°, find ∠BPQ.

Solution :

In ΔABC, we have

AB = AC

⇒ ∠ACB = ∠ABC, (The angles opposite to equal sides of a triangle are equal)

∠RAB = ∠ACB + ∠ABC, (By theorem 6.8)

⇒ 124° = ∠ABC + ∠ABC [∵ ∠ACB = ∠ABC]

⇒ 124° = 2∠ABC

⇒ \(\frac {124°}{2}\) = ∠ABC

⇒ ∠ABC = 62°

∴ ∠ACB = ∠ABC = 62°

Now in ΔBPC, we have

BP = BC, (Given)

∠BPC = ∠PCB, (The angles opposite to equal sides of a triangle are equal)

⇒ ∠BPC = ∠PCB = 62°, (∵ ∠PCB = ∠ACB)

In ΔBPC, we have

∠BPC + ∠PCB + ∠PBC = 180°,

(Sum of angles of a triangle = 180°)

⇒ 62° + 62° + ∠PBC = 180°

⇒ 124° + ∠PBC = 180°

⇒ ∠PBC = 180° – 124°

⇒ ∠PBC = 56°

Since, PQ || BC (Given)

∴ ∠BPQ = ∠PBC (A pair of alternate interior angles are equal)

⇒ ∠BPQ = 56°

Hence, ∠BPQ = 56°

Question 12.

In figure, ∠B > ∠C and L is a point on BC such that AL is the bisector of ∠BAC. If AM ⊥ BC, prove that ∠LAM = \(\frac {1}{2}\)(∠B – ∠C). [NCERT Exemplar Problems]

Solution :

Given : In ΔABC, AL is the bisector of ∠BAC and AM ⊥ BC.

To prove : ∠LAM = \(\frac {1}{2}\)(∠B – ∠C).

Proof : Since AL is the bisector of ∠BAC.

∴ ∠BAL = ∠CAL

⇒ ∠1 + ∠2 = ∠3 ………….(i)

In ΔAMB, ∠AMB = 90°, [∵ AM ⊥ BC]

∴ ∠ABM + ∠AMB + ∠1 = 180°,

(∵ Sum of angles of a triangle = 180°)

⇒ ∠ABM + 90° + ∠1 = 180°

⇒ ∠ABM = 180° – 90° – ∠1

⇒ ∠ABM = 90° – ∠1 ……..(ii)

Again in ΔAMC, we have

∠ACM + ∠AMC + ∠2 + ∠3 = 180°,

(∵ Sum of angles of a triangle = 180°)

⇒ ∠ACM + 90° + ∠2 + ∠3 = 180°

⇒ ∠ACM = 180° – 90° – ∠2 – ∠3

⇒ ∠ACM = 90° – ∠2 – ∠3 …….(iii)

Subtracting (iii) from (ii), we get

∠ABM – ∠ACM = 90° – ∠1 – 90° + ∠2 + ∠3

⇒ ∠B – ∠C = ∠2 + ∠3 – ∠1

⇒ ∠B – ∠C = ∠2 + (∠1 + ∠2) – ∠1,

[From (i), ∠3 = ∠1 + ∠2]

⇒ ∠B – ∠C = 2∠2

⇒ ∠2 = \(\frac {1}{2}\)(∠B – ∠C)

⇒ ∠LAM = \(\frac {1}{2}\)(∠B – ∠C).

Hence proved.

Question 13.

In figure, P and Q are two plane mirrors perpendicular to each other. Show that incident ray AB is parallel to the reflected ray CD.

Solution :

∵ Angle of incidence = Angle of reflection

∴ ∠1 = ∠2 and ∠3 = ∠4

∴ ∠ABC = ∠1 + ∠2

= ∠2 + ∠2 = 2∠2 …(i)

and ∠DCB = ∠3 + ∠4 = ∠3 + ∠3

= 2∠3 …(ii)

Since, the two mirrors are perpendicular, the rays perpendicular to them are also perpendicular.

i.e., BO ⊥ OC or ∠BOC = 90°

In ΔBOC, we have

∠2 + ∠3 + ∠BOC= 180°,

(Sum of angles of a triangle = 180°)

⇒ ∠2 + ∠3 + 90° = 180°

⇒ ∠2 + ∠3 = 180° – 90°

⇒ ∠2 + ∠3 = 90°

⇒ 2∠2 + 2∠3 = 180°

⇒ ∠ABC + ∠DCB = 180°, [Using (i) and (ii)]

Thus, sum of a pair of co-interior angles ∠ABC and ∠DCB is 180°.

By theorem 6.5, we have

AB || CD. Hence proved

Multiple Choice Questions

Choose the correct alternative each of the following :

Question 1.

If two angles are supplementary of each other, then each angle is :

(a) a right angle

(b) an acute angle

(c) an obtuse angle

(d) a straight angle

Solution :

(a) a right angle

Question 2.

The complement of 47° 32′ 45″ is:

(a) 43° 28′ 15″

(b) 43° 27′ 15″

(c) 42° 27′ 15″

(d) 43° 27′ 15″

Solution :

(c) 42° 27′ 15″

Question 3.

The supplement of 75° 30′ 50″ is :

(a) 14° 29′ 10″

(b) 15° 30′ 10″

(c) 105° 30′ 10″

(d) 104° 29′ 10″

Solution :

(d) 104° 29′ 10″

Question 4.

The angles a triangle are in the ratio 5 : 3 : 7 The triangle is: [NCERT Exemplar Problems]

(a) an acute angled triangle

(b) an obtused angled triangle

(c) a right triangle

(d) an isosceles triangle

Solution :

(a) an acute angled triangle

Question 5.

If angles of a triangle are in the ratio 1 : 2 : 3, then the smallest angle of the triangle is :

(a) 60°

(b) 30°

(c) 90°

(d) 50°

Solution :

(b) 30°

Question 6.

If one angle of a triangle is equal to the sum of the other two angles, then triangle is : [NCERT Exemplar Problems]

(a) a right angled triangle

(b) acute angled triangle

(c) obtuse angled triangle

(d) equilateral triangle

Solution :

(a) a right angled triangle

Question 7.

If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be : [NCERT Exemplar Problems]

(a) 50°

(b) 65°

(c) 145°

(d) 155°

Solution :

(d) 155°

Question 8.

In ΔPQR, side QR is produced to S such that ∠PRS = 110, if ∠P = ∠Q, then ∠Q is equal to :

(a) 70°

(b) 55°

(c) 50°

(d) 45°

Solution :

(b) 55°

Question 9.

In the figure, ∠x + ∠y + ∠z is equal to :

(a) 180°

(b) 90°

(c) 360°

(d) 270°

Solution :

(c) 360°

Question 10.

In the figure, ∠A + ∠B + ∠C + ∠P + ∠Q + ∠R is equal to :

(a) 180°

(b) 270°

(c) 90°

(d) 360°

Solution :

(d) 360°