RD Sharma 2019 Solutions for Class 7 Math Chapter 14 - Lines And Angles

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RD Sharma 2019 Solutions for Class 7 Math Chapter 14 Lines And Angles are provided here with simple step-by-step explanations. These solutions for Lines And Angles are extremely popular among class 7 students for Math Lines And Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2019 Book of class 7 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2019 Solutions. All RD Sharma 2019 Solutions for class 7 Math are prepared by experts and are 100% accurate.

Page No 14.10:

Question 31:

In Fig., three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.

Answer:

$∠$ BOD + $∠$ DOF + $∠$ FOA = 180°        (Linear pair)
∴ $∠$ FOA = $∠$ u = $180 ° - 90 ° - 50 ° = 40 °$
$∠ FOA = ∠ x = 40 °$     (Vertically opposite angles)
$∠ BOD = ∠ z = 90 °$     (Vertically opposite angles)
$∠ EOC = ∠ y = 50 °$     (Vertically opposite angles)

Page No 14.10:

Question 32:

In Fig., find the values of x, y and z.

Answer:

$∠ y = 25 ° (Vertically opposite angles) Since ∠ x + ∠ y = 180 ° (Linear pair) ∴ ∠ x = 180 ° - 25 ° = 155 ° ∠ z = ∠ x = 155 ° (Vertically opposite angles)$

Page No 14.20:

Question 1:

In Fig., line n is a transversal to lines l and m. Identify the following:

(i) Alternate and corresponding angles in Fig. (i).
(ii) Angles alternate to ∠d and ∠g and angles corresponding to angles ∠f and ∠h in Fig. (ii).
(iii) Angle alternate to ∠PQR, angle corresponding to ∠RQF and angle alternate to ∠PQE in Fig. (iii).
(iv) Pairs of interior and exterior angles on the same side of the transversal in Fig. (ii).

Answer:

(i) Figure (i)
Corresponding angles:
$∠$ EGB and $∠$ GHD
$∠$ HGB and $∠$ FHD
$∠$ EGA and $∠$ GHC
$∠$ AGH and $∠$ CHF
Alternate angles:
$∠$ EGB and $∠$ CHF
$∠$ HGB and $∠$ CHG
$∠$ EGA and $∠$ FHD
$∠$ AGH and $∠$ GHD

(ii) Figure (ii)
Alternate angle to $∠$ d is $∠$ e.
Alternate angle to $∠$ g is $∠$ b.
Also,
Corresponding angle to $∠$ f is $∠$ c.
Corresponding angle to $∠$ h is $∠$ a.

(iii) Figure (iii)
Angle alternate to $∠$ PQR is $∠$ QRA.
Angle corresponding to $∠$ RQF is $∠$ ARB.
Angle alternate to $∠$ POE is $∠$ ARB.

(iv) Figure (ii)
Pair of interior angles are
$∠$ a and $∠$ e
$∠$ d and $∠$ f
Pair of exterior angles are
$∠$ b and $∠$ h
$∠$ c and $∠$ g

Page No 14.20:

Question 2:

In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠CMQ = 60°, find all other angles in the figure.

Answer:

$∠$ ALM = $∠$ CMQ = $60 °$         (Corresponding angles)
$∠$ LMD = $∠$ CMQ = $60 °$         (Vertically opposite angles)
$∠$ ALM = $∠$ PLB = $60 °$           (Vertically opposite angles)
Since
$∠$ CMQ + $∠$ QMD = $180 °$      (Linear pair)
$∴ ∠$ QMD = $180 ° - 60 ° = 120 °$
$∠$ QMD = $∠$ MLB = $120 °$         (Corresponding angles)
$∠$ QMD = $∠$ CML = $120 °$         (Vertically opposite angles)
$∠$ MLB = $∠$ ALP = $120 °$         (Vertically opposite angles)

Page No 14.20:

Question 3:

In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.

Answer:

In the given Fig., AB || CD.
$∠ ALM = ∠ LMD = 35 ° (Alternate interior angles) Since ∠ PLA + ∠ ALM = 180 ° (Linear pair) ∴ ∠ PLA = 180 ° - 35 ° = 145 °$

Page No 14.20:

Question 4:

The line n is transversal to line l and m in Fig. Identify the angle alternate to ∠13, angle corresponding to ∠15, and angle alternate to ∠15.

Answer:

In this given Fig., line l || m.
Here,
Alternate angle to $∠$ 13 is $∠$ 7.
Corresponding angle to $∠$ 15 is $∠$ 7.
Alternate angle to $∠$ 15 is $∠$ 5.

Page No 14.21:

Question 5:

In Fig., line l || m and n is a transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.

Answer:

In the given figure, l || m.
Here,
$∠ 1 + ∠ 2 = 180 ° (Linear pair) ∴ ∠ 2 = 180 ° - ∠ 1 = 180 ° - 40 ° = 140 ° ∠ 5 = ∠ 1 = 40 ° (Corresponding angles) ∠ 3 = ∠ 1 = 40 ° (Vertically opposite angles) ∠ 7 = ∠ 3 = 40 ° (Corresponding angles) ∠ 7 = ∠ 5 = 40 ° (Vertically opposite angles)$
Also,
$∠ 2 = ∠ 6 = 140 ° (Corresponding angles) ∠ 2 = ∠ 4 = 140 ° (Vertically opposite angles) ∠ 4 = ∠ 8 = 140 ° (Corresponding angles) ∠ 8 = ∠ 6 = 40 ° (Vertically opposite angles)$
Thus,
$∠ 2 = ∠ 8, ∠ 3 = ∠ 5, ∠ 6 = ∠ 4, ∠ 1 = ∠ 7$
Hence, alternate angles are equal.

Page No 14.21:

Question 6:

In Fig., line l || m and a transversal n cuts them at P and Q respectively. If ∠1 = 75°, find all other angles.

Answer:

In the given figure, l || m, n is a transversal line and ∠1 = 75°.
Thus, we have:
$∠ 1 + ∠ 2 = 180 ° (Linear pair) \Rightarrow ∠ 2 = 180 ° - ∠ 1 = 180 ° - 75 ° = 105 ° ∴ ∠ 1 = ∠ 5 = 75 ° (Corresponding angles) ∠ 1 = ∠ 3 = 75 ° (Vertically opposite angles) ∠ 5 = ∠ 7 = 75 ° (Vertically opposite angles) Now, ∠ 2 = ∠ 6 = 105 ° (Corresponding angles) ∠ 6 = ∠ 8 = 105 ° (Vertically opposite angles) ∠ 2 = ∠ 4 = 105 ° (Vertically opposite angles)$

Page No 14.21:

Question 7:

In Fig., AB || CD and a transversal PQ cuts them at L and M respectively. If ∠QMD = 100°, find all other angles.

Answer:

In the given figure, AB || CD, PQ is a transversal line and $∠$ QMD = 100°.
Thus, we have:
$∠$ DMQ + $∠$ QMC = 180°    (Linear pair)
$∴ ∠ QMC = 180 ° - ∠ DMQ = 180 ° - 100 ° = 80 °$
Thus,
$∠$ DMQ = $∠$ BLM = 100°         (Corresponding angles)
$∠$ DMQ = $∠$ CML = 100°         (Vertically opposite angles)
$∠$ BLM = $∠$ PLA = 100°           (Vertically opposite angles)
Also,
$∠$ CMQ = $∠$ ALM = 80°         (Corresponding angles)
$∠$ CMQ = $∠$ DML = 80°         (Vertically opposite angles)
$∠$ ALM = $∠$ PLB = 80°           (Vertically opposite angles)

Page No 14.21:

Question 8:

In Fig., l || m and p || q. Find the values of x, y, z, t.

Answer:

In the given figure, l || m and p || q.
Thus, we have:
$∠ z = 80 °$                 (Vertically opposite angles)
$∠ z = ∠ t = 80 °$        (Corresponding angles)
$∠ z = ∠ y = 80 °$        (Corresponding angles)
$∠ x = ∠ y = 80 °$        (Corresponding angles)

Page No 14.21:

Question 9:

In Fig., line l || m, ∠1 = 120° and ∠2 = 100°, find out ∠3 and ∠4.

Answer:

In the given figure, ∠1 = 120° and ∠2 =100°.
Since l || m, so
$∠ 2 = ∠ 5 = 100 ° (Alternate interior angles) ∠ 5 + ∠ 3 = 180 ° (Linear pair) \Rightarrow ∠ 3 = 180 ° - ∠ 5 = 180 ° - 100 ° = 80 °$
Also,
$∠ 1 + ∠ 6 = 180 ° (Linear pair) \Rightarrow ∠ 6 = 180 ° - ∠ 1 = 180 ° - 120 ° = 60 °$
We know that the sum of all the angles of triangle is 180°.
$∴ ∠ 6 + ∠ 3 + ∠ 4 = 180 ° \Rightarrow 60 ° + 80 ° + ∠ 4 = 180 ° \Rightarrow 140 ° + ∠ 4 = 180 ° \Rightarrow ∠ 4 = 180 ° - 140 ° = 40 °$

Page No 14.21:

Question 10:

In Fig., line l || m. Find the values of a, b, c, d. Give reasons.

Answer:

In the given figure, line l || m.
Thus, we have:
$∠ a = 110 ° (Vertically opposite angles) ∠ b = ∠ a = 110 ° (Corresponding angles) ∠ d = 85 ° (Vertically opposite angles) ∠ c = ∠ d = 85 ° (Corresponding angles)$

Page No 14.22:

Question 11:

In Fig., AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.

Answer:

In the given figure, AB || CD and t is a transversal line.
Now, let:
$∠ 1 = 3 x ∠ 2 = 2 x$
Thus, we have:
$∠ 1 + ∠ 2 = 180 ° (Linear pair) ∴ 3 x + 2 x = 180 ° \Rightarrow 5 x = 180 ° \Rightarrow x = \frac{180 °}{5} = 36 ° Thus, ∠ 1 = 3 \times 36 ° = 108 ° ∠ 2 = 2 \times 36 ° = 72 °$
Now,
$∠ 1 = ∠ 5 = 108 ° (Corresponding angles) ∠ 1 = ∠ 3 = 108 ° (Vertically opposite angles) ∠ 5 = ∠ 7 = 108 ° (Vertically opposite angles) ∠ 2 = ∠ 6 = 72 ° (Corresponding angles) ∠ 4 = ∠ 2 = 72 ° (Vertically opposite angles) ∠ 8 = ∠ 6 = 72 ° (Vertically opposite angles)$

Page No 14.22:

Question 12:

In Fig., l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠1, ∠2 and ∠3.

Answer:

In the given figure, l || m || n and p is a transversal line.
Thus, we have:
$∠ 4 + 60 ° = 180 ° (Linear pair) \Rightarrow ∠ 4 = 180 ° - 60 ° = 120 ° ∠ 4 = ∠ 1 = 120 ° (Corresponding angles) ∠ 1 = ∠ 2 = 120 ° (Corresponding angles) ∠ 3 = ∠ 2 = 120 ° (Vertically opposite angles) Thus, ∠ 1 = ∠ 2 = ∠ 3 = 120 °$

Page No 14.22:

Question 13:

In Fig., if l || m || n and ∠1 = 60°, find ∠2.

Answer:

In the given figure, l || m || n and ∠1 = 60°.
Thus, we have:
$∠ 3 = ∠ 1 = 60 ° (Corresponding angle) Now, ∠ 3 + ∠ 4 = 180 ° (Linear pair) ∠ 4 = 180 ° - ∠ 3 = 180 ° - 60 ° = 120 ° ∠ 2 = ∠ 4 = 120 ° (Alternate interior angles)$

Page No 14.22:

Question 14:

In Fig., if AB || CD and CD || EF, find ∠ACE.

Answer:

In the given figure, AB || CD and CD || EF.
Extend line CE to E^'.

Thus, we have:
$∠ BAC = ∠ ACD = 70 ° (Alternate angles) Now, ∠ 3 + ∠ CEF = 180 ° (Linear pair) \Rightarrow ∠ 3 = 180 ° - ∠ CEF = 180 ° - 130 ° = 50 ° Since CD | | EF, then ∠ 2 = ∠ 3 = 50 ° (Corresponding angles) ∠ ACE = ∠ ACD - ∠ 2 = 70 ° - 50 ° = 20 °$

Page No 14.22:

Question 15:

In Fig., if l || m, n || p and ∠1 = 85°, find ∠2.

Answer:

In the given figure, l || m, n || p and ∠1 = 85°.
Now, let ∠4 be the adjacent angle of ∠2.
Thus, we have:
$∠ 3 = ∠ 1 = 85 ° (Corresponding angles)$
$∠ 3 + ∠ 2 = 180 °$ (Sum of interior angles on the same side of the transversal)
$∴ ∠ 2 = 180 ° - ∠ 3 = 180 ° - 85 ° = 95 °$

Page No 14.22:

Question 16:

In Fig., a transversal n cuts two lines l and m. If ∠1 = 70° and ∠7 = 80°, is l || m?

Answer:

We know that if the alternate exterior angles of two lines are equal, then the lines are parallel.
In the given figure, $∠ 1 and ∠ 7$ are alternate exterior angles, but they are not equal.
$∠ 1 \neq ∠ 7 70 ° \neq 80 °$

Therefore, lines l and m are not parallel.

Page No 14.23:

Question 17:

In Fig., a transversal n cuts two lines l and m such that ∠2 = 65° and ∠8 = 65°. Are the lines parallel?

Answer:

$∠$ 2 = $∠$ 3 = 65°    (Vertically opposite angles)
$∠$ 8 = $∠$ 6 = 65°    (Vertically opposite angles)
∴ $∠$ 3 = $∠$ 6
⇒ l || m         (Two lines are parallel if the alternate angles formed with the transversal are equal)

Page No 14.23:

Question 18:

In Fig., show that AB || EF.

Answer:

Extend line CE to E'.

$∠ BAC = 57 ° = 22 ° + 35 ° = ∠ ACE + ∠ ECD ∴ AB | | CD Here, ∠ E' EF + ∠ FEC = 180 ° (Linear pair) \Rightarrow ∠ E' EF = 180 ° - ∠ FEC = 180 ° - 145 ° = 35 ° = ∠ ECD ∴ EF | | CD Thus, AB | | CD | | EF$

Page No 14.23:

Question 19:

In Fig., AB || CD. Find the values of x, y, z.

Answer:

$∠ x + 125 ° = 180 °$              (Linear pair)
$∴ ∠ x = 180 ° - 125 ° = 55 °$

$∠ z = 125 °$             (Corresponding angles)
$∠ x + ∠ z = 180 °$    (Sum of adjacent interior angles is $180 °$ )
$∠ x + 125 ° = 180 ° \Rightarrow ∠ x = 180 ° - 125 ° = 55 °$

$∠ x + ∠ y = 180 °$    (Sum of adjacent interior angles is $180 °$ )
$55 ° + ∠ y = 180 ° \Rightarrow ∠ y = 180 ° - 55 ° = 125 °$

Page No 14.23:

Question 20:

In Fig., find out ∠PXR, if PQ || RS.

Answer:

Draw a line parallel to PQ passing through X.

Here,
$∠ PQX = ∠ PXF = 70 ° and ∠ SRX = ∠ RXF = 50 °$ (Alternate interior angles)
âˆµ PQ || RS || XF
∴ $∠ PXR = ∠ PXF + ∠ FXR = 70 ° + 50 ° = 120 °$

Page No 14.23:

Question 21:

In Fig., we have

(i) ∠MLY = 2 ∠LMQ, find ∠LMQ.
(ii) ∠XLM = (2x − 10)° and ∠LMQ = x + 30°, find x.
(iii) ∠XLM = ∠PML, find ∠ALY
(iv) ∠ALY = (2x − 15)°, and ∠LMQ = (x + 40)°, find x

Answer:

(i)
$∠ LMQ = ∠ ALY (Corresponding angles) ∴ ∠ MLY + ∠ ALY = 180 ° (Linear pair) \Rightarrow 2 ∠ ALY + ∠ ALY = 180 ° \Rightarrow 3 ∠ ALY = 180 ° \Rightarrow ∠ ALY = \frac{180 °}{3} = 60 ° ∴ ∠ LMQ = 60 °$

(ii)
$∠ XLM = ∠ LMQ (Alternate interior angles) \Rightarrow (2 x - 10) ° = (x + 30) ° \Rightarrow 2 x - x = 30 ° + 10 ° \Rightarrow x = 40 °$

(iii)
$∠ ALX = ∠ LMP (Corresponding angles) ∠ ALX + ∠ XLM = 180 ° (Linear pair) ∠ XLM = ∠ LMP (Given) ∴ ∠ LMP + ∠ LMP = 180 ° \Rightarrow 2 ∠ LMP = 180 ° \Rightarrow ∠ LMP = \frac{180 °}{2} = 90 ° ∠ XLM = ∠ LMP = 90 ° ∠ ALY = ∠ XLM (Vertically opposite angles) ∴ ∠ ALY = 90 °$

(iv)
$∠ ALY = ∠ LMQ (Corresponding angles) ∴ (2 x - 15) ° = (x + 40) ° \Rightarrow 2 x - x = 40 ° + 15 ° \Rightarrow x = 55 °$

Page No 14.23:

Question 22:

In Fig., DE || BC. Find the values of x and y.

Answer:

$∠$ ABC = $∠$ DAB (Alternate interior angles)
$∴ x = 40 °$

$∠$ ACB = $∠$ EAC (Alternate interior angles)
$∴ y = 55 °$

Page No 14.24:

Question 23:

In Fig., line AC || line DE and ∠ABD = 32°. Find out the angles x and y if ∠E = 122°.

Answer:

$∠ BDE = ∠ ABD = 32 ° (Alternate interior angles) \Rightarrow ∠ BDE + y = 180 ° (Linear pair) \Rightarrow 32 ° + y = 180 ° \Rightarrow y = 180 ° - 32 ° = 148 °$

$∠ ABE = ∠ E = 122 ° (Alternate interior angle) ∠ ABD + ∠ DBE = 122 ° 32 ° + x = 122 ° x = 122 ° - 32 ° = 90 °$

Page No 14.24:

Question 24:

In Fig., side BC of âˆ†ABC has been produced to D and CE || BA. If ∠ABC = 65°, ∠BAC = 55°, find ∠ACE, ∠ECD and ∠ACD.

Answer:

$∠$ ABC = $∠$ ECD = 55° (Corresponding angles)
$∠$ BAC = $∠$ ACE = 65° (Alternate interior angles)
Now, $∠$ ACD = $∠$ ACE + $∠$ ECD
⇒ $∠$ ACD = 55° + 65° = 120°

Page No 14.24:

Question 25:

In Fig., line CA ⊥ AB || line CR and line PR || line BD. Find ∠x, ∠y and ∠z.

Answer:

Since CA ⊥ AB,
$∴ ∠ x = 90 °$
We know that the sum of all the angles of triangle is 180°.
$In ∆ APQ, ∠ QAP + ∠ APQ + ∠ PQA = 180 ° \Rightarrow 90 ° + ∠ APQ + 20 ° = 180 ° \Rightarrow 110 ° + ∠ APQ = 180 ° \Rightarrow ∠ APQ = 180 ° - 110 ° = 70 °$
$∠$ PBC = $∠$ APQ = $70 °$ (Corresponding angles)
Since $∠ PRC + ∠ z = 180 ° (Linear pair)$
$∴ ∠ z = 180 ° - 70 ° = 110 ° [∠ APQ = ∠ PRC (Alternate interior angles)]$

Page No 14.24:

Question 26:

In Fig., PQ || RS. Find the value of x.

Answer:

$∠ RCD + ∠ RCB = 180 ° (Linear pair) \Rightarrow ∠ RCB = 180 ° - 130 ° = 50 ° In △ ABC, ∠ BAC + ∠ ABC + ∠ BCA = 180 ° (Angle sum property) \Rightarrow ∠ BAC = 180 ° - 55 ° - 50 ° = 75 °$

Page No 14.24:

Question 27:

In Fig., AB || CD and AE || CF; ∠FCG = 90° and ∠BAC = 120°. Find the values of x, y and z.

Answer:

$∠$ BAC = $∠$ ACG = 120°    (Alternate interior angle)
∴ $∠$ ACF + $∠$ FCG = 120°
⇒ $∠$ ACF = 120° − 90° = 30°

$∠$ DCA + $∠$ ACG = 180°            (Linear pair)
⇒ $∠$ x = 180° − 120° = 60°

$∠$ BAC + $∠$ BAE + $∠$ EAC = 360°
$∠$ CAE = 360° − 120° − (60° + 30°) = 150°             ( $∠$ BAE = $∠$ DCF)

Page No 14.25:

Question 28:

In Fig., AB || CD and AC || BD. Find the values of x, y, z.

Answer:

(i) Since AC || BD and CD || AB, ABCD is a parallelogram.
$∠$ CAB + $∠$ ACD = 180°     (Sum of adjacent angles of a parallelogram)
∴ $∠$ ACD = 180° − 65° = 115°
$∠$ CAD = $∠$ CDB = 65° (Opposite angles of a parallelogram)
$∠$ ACD = $∠$ DBA = 115° (Opposite angles of a parallelogram)

(ii) Here,
AC || BD and CD || AB
$∠$ DAC = x = 40°      (Alternate interior angle)
$∠$ DAB = y = 35°      (Alternate interior angle)

Page No 14.25:

Question 29:

In Fig., state which lines are parallel and why?

Answer:

Let F be the point of intersection of line CD and the line passing through point E.

Since $∠$ ACD and $∠$ CDE are alternate and equal angles, so
$∠$ ACD = 100° = $∠$ CDE
∴ AC || EF

Page No 14.25:

Question 30:

In Fig. 87, the corresponding arms of ∠ABC and ∠DEF are parallel. If ∠ABC = 75°, find ∠DEF.

Answer:

Construction: Let G be the point of intersection of lines BC and DE.

âˆµ AB || DE and BC || EF

∴ $∠ ABC = ∠ DGC = ∠ DEF = 75 °$ (Corresponding angles)â€‹

Page No 14.26:

Question 1:

The sum of an angle and one third of its supplementary angle is 90°. The measure of the angle is
(a) 135°
(b) 120°
(c) 60°
(d) 45°

Answer:

Let the required angle be x.
Now, supplementary of the required angle = 180^âˆ˜− x
Then,
$x + \frac{1}{3} (180 ° - x) = 90 ° \Rightarrow 3 x + 180 ° - x = 270 ° \Rightarrow 2 x = 90 ° \Rightarrow x = 45 °$
Hence, the correct answer is option (d).

Page No 14.26:

Question 2:

If angles of a linear pair are equal, then the measure of each angle is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

Let the required angle be x
Now, Sum of linear pair angles = 180^âˆ˜
⇒ x + x = 180^âˆ˜
⇒ 2x = 180^âˆ˜
⇒ x = 90^âˆ˜
Hence, the correct answer is option (d).

Page No 14.26:

Question 3:

Two complemntary angles are in the ratio 2 : 3. The measure of the larger angle is
(a) 60°
(b) 54°
(c) 66°
(d) 48°

Answer:

Let the angles be 2x and 3x.
Now, 2x + 3x = 90^âˆ˜
⇒ 5x = 90^âˆ˜
⇒ x = 18^âˆ˜
∴ Larger angle = 3x = 3 × 18^âˆ˜ = 54^âˆ˜
Hence, the correct answer is option (b).

Page No 14.26:

Question 4:

An angle is thrice its supplement. The measure of the angle is
(a) 120°
(b) 105°
(c) 135°
(d) 150°

Answer:

Let the required angle be x.
Then,
$x = 3 (180 ° - x) \Rightarrow x = 540 ° - 3 x \Rightarrow 4 x = 540 ° \Rightarrow x = 135 °$
Hence, the correct answer is option (c).

Page No 14.26:

Question 5:

In Fig. 88 PR is a straight line and ∠PQS : ∠SQR = 7 : 5. The measure of ∠SQR is
(a) 60°
(b) ${(62 \frac{1}{2})}^{°}$
(c) ${(67 \frac{1}{2})}^{°}$
(d) 75°

Answer:

Let the measures of the angle ∠PQS and ∠SQR be 7x and 5x.
Now, ∠PQS + ∠SQR = 180^âˆ˜ [Linear pair angles]
⇒ 7x + 5x = 180^âˆ˜
⇒ 12x = 180^âˆ˜
⇒ x = 15^âˆ˜
∴ ∠SQR = 5x = 5 × 15^âˆ˜ = 75^âˆ˜
Hence, the correct answer is option (d).

Page No 14.26:

Question 6:

The sum of an angle and half of its complementary angle is 75°. The measure of the angle is
(a) 40°
(b) 50°
(c) 60°
(d) 80°

Answer:

Let the required angle be x.
Now, complementnary of the required angle = 90^âˆ˜− x
Then,
$x + \frac{1}{2} (90 ° - x) = 75 ° \Rightarrow 2 x + 90 ° - x = 150 ° \Rightarrow x = 150 - 90 ° \Rightarrow x = 60 °$
Hence, the correct answer is option (c).

Page No 14.26:

Question 7:

∠A is an obtuse angle. The measure of ∠A and twice its supplementary differ by 30°. Then ∠A can be
(a) 150°
(b) 110°
(c) 140°
(d) 120°

Answer:

Supplementary of ∠A = 180^âˆ˜ − ∠A
Now,
∠A + 30^âˆ˜ = 2(180^âˆ˜ − ∠A)
⇒ ∠A + 30^âˆ˜ = 360^âˆ˜ − 2∠A
⇒ 3∠A = 360^âˆ˜ − 30^âˆ˜
⇒ 3∠A = 330^âˆ˜
⇒ ∠A = 110^âˆ˜
Hence, the correct answer is option (b).

Page No 14.26:

Question 8:

An angle is double of its supplement. The measure of the angle is
(a) 60°
(b) 120°
(c) 40°
(d) 80°

Answer:

Let the required angle be x.
Now, supplementary of the required angle = 180^âˆ˜− x
Then,
$x = 2 (180 ° - x) \Rightarrow x = 360 ° - 2 x \Rightarrow 3 x = 360 ° \Rightarrow x = 120 °$
Hence, the correct answer is option (b).

Page No 14.26:

Question 9:

The measure of an angle which is its own complement is
(a) 30°
(b) 60°
(c) 90°
(d) 45°

Answer:

Let the required angle be x.
Now, complementary of the required angle = 90^âˆ˜− x
Then,
$x = 90 ° - x \Rightarrow x = 90 ° - x \Rightarrow 2 x = 90 ° \Rightarrow x = 45 °$
Hence, the correct answer is option (d).

Page No 14.26:

Question 10:

Two supplementary angles are in the ratio 3 : 2. The smaller angle measures
(a) 108°
(b) 81°
(c) 72°
(d) 68°

Answer:

Let the angles be 3x and 2x.
Now, 3x + 2x = 180^âˆ˜
⇒ 5x = 180^âˆ˜
⇒ x = 36^âˆ˜
∴ Smaller angle = 2x = 2 × 36^âˆ˜ = 72^âˆ˜
Hence, the correct answer is option (c).

Page No 14.26:

Question 11:

In Fig. 89, the value of x is
(a) 75
(b) 65
(c) 45
(d) 55

Answer:

∠AOC and ∠BOC = 180^âˆ˜ [âˆµ Linear pair angles]
⇒ 44^âˆ˜+ (2x + 6)^âˆ˜ = 180^âˆ˜
⇒ (2x + 6)^âˆ˜ = 136^âˆ˜
⇒ 2x + 6 = 136
⇒ 2x = 130
⇒ x = 65
Hence, the correct answer is option (b).

Page No 14.27:

Question 12:

In Fig. 90, AOB is a straight line and the ray OCstands on it. The value of x is
(a) 16
(b) 26
(c) 36
(d) 46

Answer:

∠AOC + ∠BOC = 180^âˆ˜ [âˆµ Linear pair angles]
⇒ (2x + 15)^âˆ˜ + (3x + 35)^âˆ˜ = 180^âˆ˜
⇒ (5x + 50)^âˆ˜ = 180^âˆ˜
⇒ 5x + 50 = 180
⇒ 5x = 130
⇒ x = 26
Hence, the correct answer is option (b).

Page No 14.27:

Question 13:

In Fig. 91, AOB is a straight line and 4x = 5y. The value of x is
(a) 100
(b) 105
(c) 110
(d) 115

Answer:

∠AOC + ∠BOC = 180^âˆ˜ [âˆµ Linear pair angles]
⇒ y^âˆ˜ + x^âˆ˜ = 180^âˆ˜
⇒ y + x = 180
$\Rightarrow \frac{4 x}{5} + x = 180 [∵ 4 x = 5 y \Rightarrow y = \frac{4 x}{5}] \Rightarrow 4 x + 5 x = 180 \times 5 \Rightarrow 9 x = 180 \times 5 \Rightarrow x = 100$
Hence, the correct answer is option (a).

Page No 14.27:

Question 14:

In Fig. 92, AOB is a straight line such that ∠AOC = (3x + 10)°, ∠COD = 50° and ∠BOD = (x − 8)°. The value of x is
(a) 32
(b) 36
(c) 42
(d) 52

Answer:

∠AOC + ∠COD + ∠BOD = 180^âˆ˜ [AOB is a straight line]
⇒ (3x + 10)^âˆ˜ + 50^âˆ˜+ (x − 8)^âˆ˜ = 180^âˆ˜
⇒ 3x + 10 + 50+ x − 8 = 180
⇒ 4x + 52 = 180
⇒ 4x = 128
⇒ x = 32
Hence, the correct answer is option (a).

Page No 14.27:

Question 15:

In Fig. 93, if AOC is a straight line, then x =
(a) 42°
(b) 52°
(c) 142°
(d) 38°

Answer:

∠AOD + ∠DOB + ∠BOC = 180^âˆ˜ [âˆµ AOC is a straight line]
⇒ 38^âˆ˜ + x+ 90^âˆ˜ = 180^âˆ˜
⇒ x + 128^âˆ˜ = 180^âˆ˜
⇒ x = 52^âˆ˜
Hence, the correct answer is option (b).

Page No 14.28:

Question 16:

In Fig. 94, if ∠AOC is a straight line, then the value of x is
(a) 15
(b) 18
(c) 20
(d) 16

Answer:

∠AOD + ∠DOB + ∠BOC = 180^âˆ˜ [ AOC is a straight line]
⇒ 2x^âˆ˜ + 90^âˆ˜ + 3x^âˆ˜= 180^âˆ˜
⇒ 5x^âˆ˜ + 90^âˆ˜ = 180^âˆ˜
⇒ 5x = 90
⇒ x = 18
Hence, the correct answer is option (b).

Page No 14.28:

Question 17:

In Fig. 95, if AB, CD and EF are straight lines, then x =
(a) 5
(b) 10
(c) 20
(d) 30

Answer:

Let all the lines intersect at O.

∠COF = ∠DOE = 4x^âˆ˜ [Vertically opposite angles]
∠AOC + ∠COF + ∠BOF = 180^âˆ˜ [AOB is a straight line]
⇒ 2x^âˆ˜ + 4x^âˆ˜ + 3x^âˆ˜= 180^âˆ˜
⇒ 9x^âˆ˜ = 180^âˆ˜
⇒ 9x = 180
⇒ x = 20
Hence, the correct answer is option (c).

Page No 14.28:

Question 18:

In Fig. 96, if AB, CD and EF are straight lines, then x + y + z =
(a) 180
(b) 203
(c) 213
(d) 134

Answer:

∠DAE + ∠BAD + ∠BAF = 180^âˆ˜ [EAF is a straight line]
⇒ 3x^âˆ˜ + 49^âˆ˜ + 62^âˆ˜= 180^âˆ˜
⇒ 3x^âˆ˜ + 111^âˆ˜= 180^âˆ˜
⇒ 3x^âˆ˜ = 69^âˆ˜
⇒ 3x = 69
⇒ x = 23
Now, ∠CAE + ∠CAF = 180^âˆ˜ [âˆµ EAF is a straight line]
⇒ z^âˆ˜ + y^âˆ˜= 180^âˆ˜
⇒ z + y= 180
Now, x + y + z = 23 + 180 = 203
Hence, the correct answer is option (b).

Page No 14.28:

Question 19:

In Fig. 97, if AB is parallel to CD, then the value of ∠BPE is
(a) 106°
(b) 76°
(c) 74°
(d) 84°

Answer:

Since, AB || CD
∴ ∠BPQ = ∠PQC [Alternate interior angles]
⇒ (3x + 34)^âˆ˜ = (5x − 14)^âˆ˜
⇒ 3x + 34 = 5x − 14
⇒ 48 = 2x
⇒ x = 24
∴ ∠BPQ = (3 × 24 + 34)^âˆ˜ = 106^âˆ˜
∠BPQ + ∠BPE = 180^âˆ˜ [EF is a straight line]
⇒ 106^âˆ˜ + ∠BPE = 180^âˆ˜
⇒ ∠BPE = 74^âˆ˜
Hence, the correct answer is option (c).

Page No 14.29:

Question 20:

In Fig. 98, if AB is parallel to CO and EF is a transversal, then x =
(a) 19
(b) 29
(c) 39
(d) 49

Answer:

Let the line EF intersect AB and CD at P and Q respectively.

Since, AB || CD
∴ ∠BPQ + ∠PQD = 180^âˆ˜ (Angles on the same side of a transversal line are supplementary)
⇒ (7x − 12)^âˆ˜ + (4x + 17)^âˆ˜ = 180^âˆ˜
⇒ 7x − 12 + 4x + 17 = 180
⇒ 11x + 5 = 180
⇒ 11x = 175
⇒ x = 15.90

Disclaimer: No option is correct.

Page No 14.29:

Question 21:

In Fig. 99, AB || CD and EF is a transversal intersecting ABand CO at Pand Q respectively. The measure of ∠DPQ is
(a) 100^âˆ˜
(b) 80^âˆ˜
(c) 110^âˆ˜
(d) 70^âˆ˜

Answer:

∠BQF = ∠AQP = (4x)^âˆ˜             [Vertically opposite angles]
Since, AB || CD
∴ ∠AQP + ∠CPQ = 180^âˆ˜         [Angles on the same side of a transversal line are supplementary]
⇒ (4x)^âˆ˜ + (5x)^âˆ˜ = 180^âˆ˜
⇒ 9x = 180
⇒ x = 20
∴ ∠BQF = (4 × 20)^âˆ˜ = 80^âˆ˜
Now, ∠BQF = ∠DPQ = 80^âˆ˜          [Corresponding angles]
Hence, the correct answer is option (b).

Page No 14.30:

Question 22:

In Fig. 100, AB || CO and EF is a transversal intersecting AB and CD at P and Q respective. The measure of ∠OOP is
(a) 65
(b) 25
(c) 115
(d) 105

Answer:

∠BPE = ∠APQ = (5x − 10)^âˆ˜        [Vertically opposite angles]
Since, AB || CD
∴ ∠APQ + ∠CQP = 180^âˆ˜            [Angles on the same side of a transversal line are supplementary]
⇒ (5x − 10)^âˆ˜ + (3x − 10)^âˆ˜ = 180^âˆ˜
⇒ 8x − 20 = 180
⇒ 8x = 200
⇒ x = 25
∴ ∠BPE = (5 × 25 − 10)^âˆ˜ = 115^âˆ˜
Now, ∠BPE = ∠DQP = 115^âˆ˜          [Corresponding angles]
Hence, the correct answer is option (c).

Page No 14.30:

Question 23:

In Fig. 101, AB || CD and EF is a transversal. The value of y − x is
(a) 30
(b) 35
(c) 95
(d) 25

Answer:

Since, AB || CD
∴ ∠BPQ = ∠DQF [Corresponding angles]
⇒ (5x − 20)^âˆ˜ = (3x + 40)^âˆ˜
⇒ 5x − 20 = 3x + 40
⇒ 2x = 60
⇒ x = 30
∴ ∠BPQ = (5 × 30 − 20 )^âˆ˜ = 130^âˆ˜
Now, ∠APE = ∠BPQ [Vertically opposite angles]
⇒ 2y^âˆ˜ = 130^âˆ˜
⇒ y = 65
∴ y − x = 65 − 30 = 35
Hence, the correct answer is option (b).

Page No 14.30:

Question 24:

In Fig. 102, AB || CD || EF, ∠ABG = 110°, ∠GCO = 100° and ∠BGC = x°. The value of x is
(a) 35
(b) 50
(c) 30
(d) 40

Answer:

Since, AB || EG
∴ ∠ABG + ∠EGB = 180^âˆ˜         (Angles on the same side of a transversal line are supplementary)
⇒ 110^âˆ˜ + ∠EGB = 180^âˆ˜
⇒ ∠EGB = 70^âˆ˜
Again, CD || GF
∴ ∠DCG + ∠FGC = 180^âˆ˜         (Angles on the same side of a transversal line are supplementary)
⇒ 100^âˆ˜ + ∠FGC = 180^âˆ˜
⇒ ∠FGC = 80^âˆ˜
Now, ∠EGB + ∠BGC +∠FGC = 180^âˆ˜
⇒ 70^âˆ˜ + x^âˆ˜ + 80^âˆ˜= 180^âˆ˜
⇒ 150^âˆ˜+ x^âˆ˜ = 180^âˆ˜
⇒ x^âˆ˜ = 30^âˆ˜
⇒ x = 30
Hence, the correct answer is option (c).

Page No 14.31:

Question 25:

In Fig. 103, PO || RS and ∠PAB = 60° and ∠ACS = 100°. Then, ∠BAC =
(a) 40°
(b) 60°
(c) 80°
(d) 50°

Answer:

Since, PQ || RS
∴ ∠PAC = ∠ACS = 100^âˆ˜ [Corresponding angles]
Now, ∠PAC = 100^âˆ˜
⇒ ∠PAB + ∠BAC = 100^âˆ˜
⇒ 60^âˆ˜ + ∠BAC = 100^âˆ˜
⇒ ∠BAC = 40^âˆ˜
Hence, the correct answer is option (a).

Page No 14.31:

Question 26:

In Fig. 104, AB || CO, ∠OAB = 150° and ∠OCO = 120°. Then, ∠AOC =
(a) 80°
(b) 90°
(c) 70°
(d) 100°

Answer:

Construction: Draw a line OE from the point O parallel to AB and CD

Since, AB || OE
∴ ∠BAO + ∠AOE = 180^âˆ˜ [Angles on the same side of a transversal line are supplementary]
⇒ 150^âˆ˜ + ∠AOE = 180^âˆ˜
⇒ ∠AOE = 30^âˆ˜
Again, CD || OE
∴ ∠DCO + ∠COE = 180^âˆ˜ [Angles on the same side of a transversal line are supplementary]
⇒ 120^âˆ˜ + ∠COE = 180^âˆ˜
⇒ ∠COE = 60^âˆ˜
Now, ∠AOC = ∠AOE + ∠COE
= 30^âˆ˜ + 60^âˆ˜
= 90^âˆ˜
Hence, the correct answer is option (b).

Page No 14.31:

Question 27:

In Fig. 105, if AOB and COD are straight lines. Then, x + y =
(a) 120
(b) 140
(c) 100
(d) 160

Answer:

∠AOD + ∠BOD = 180^âˆ˜ [Linear pair angles]
⇒ (7x − 20)^âˆ˜ + 3x^âˆ˜ = 180^âˆ˜
⇒ 7x − 20 + 3x = 180
⇒ 10x = 200
⇒ x = 20
∴ ∠AOD = (7 × 20 − 20)^âˆ˜ = 120^âˆ˜
Now∠AOD = ∠BOC = 120^âˆ˜ [Vertically opposite angles]
∴ y = 120
Now, x + y = 20 + 120
= 140
Hence, the correct answer is option (b).

Page No 14.31:

Question 28:

In Fig. 106, the value of x is
(a) 22
(b) 20
(c) 21
(d) 24

Answer:

(8x − 41)^âˆ˜ + (3x)^âˆ˜ + (3x + 10)^âˆ˜ + (4x − 5)^âˆ˜= 360^âˆ˜
⇒ 8x − 41 + 3x + 3x + 10 + 4x − 5 = 360
⇒ 18x − 36 = 360
⇒ 18x = 396
⇒ x = 22
Hence, the correct answer is option (a).

Page No 14.32:

Question 29:

In Fig. 107, if AOBand COD are straight lines, then
(a) x = 29, y = 100
(b) x = 110, y = 29
(c) x = 29, y = 110
(d) x = 39, y = 110

Answer:

∠AOD + ∠BOD = 180^âˆ˜         [Linear pair angles]
⇒ y^âˆ˜ + 70^âˆ˜ = 180^âˆ˜
⇒ y^âˆ˜ = 110^âˆ˜
⇒ y = 110
Now, ∠AOC = ∠BOD = 70^âˆ˜    [Vertically opposite angles]
Now, ∠AOC + ∠COE + ∠EOB + ∠BOD + ∠AOD = 360^âˆ˜            [Complete angle]
⇒ 70^âˆ˜ + 28^âˆ˜ + (3x − 5)^âˆ˜ + 70^âˆ˜ + 110^âˆ˜ = 360^âˆ˜
⇒ (3x)^âˆ˜ + 273^âˆ˜ = 360^âˆ˜
⇒ 3x = 87
⇒ x = 29
Hence, the correct answer is option (c).

Page No 14.32:

Question 30:

In Fig. 108, if AB || CD then the value of x is
(a) 87
(b) 93
(c) 147
(d) 141

Answer:

Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠FCD + Reflex∠FCD = 360^âˆ˜         (Complete angle)
⇒ ∠FCD + 273^âˆ˜ = 360^âˆ˜
⇒ ∠FCD = 87^âˆ˜
Since, PQ || CD
∴∠QFC + ∠FCD = 180^âˆ˜                 (Angles on the same side of a transversal line are supplementary)
⇒ ∠QFC + 87^âˆ˜ = 180^âˆ˜
⇒ ∠QFC = 93^âˆ˜
Now, ∠ABF = ∠BFQ              (Corresponding angles)
= ∠BFC + ∠QFC
= 54^âˆ˜ + 93^âˆ˜
= 147^âˆ˜
∴ x^âˆ˜ = 147^âˆ˜
⇒ x = 147
Hence, the correct answer is option (c).

Page No 14.32:

Question 31:

In Fig. 109, if AB || CD then the value of x is
(a) 34
(b) 124
(c) 24
(d) 158

Answer:

Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠QEC + ∠ECD = 180^âˆ˜                 [Angles on the same side of a transversal line are supplementary]
⇒ ∠QEC + 56^âˆ˜ = 180^âˆ˜
⇒ ∠QEC = 124^âˆ˜
Now, ∠BEQ + ∠QEC = ∠BEC
⇒ ∠BEQ + 124^âˆ˜ = 158^âˆ˜
⇒ ∠BEQ = 34^âˆ˜
Now, ∠ABE = ∠BEQ = 34^âˆ˜              [Corresponding angles]
∴ x^âˆ˜ = 34^âˆ˜
⇒ x = 34
Hence, the correct answer is option (a).

Page No 14.32:

Question 32:

In Fig. 110, if AB || CD. The value of x is
(a) 122
(b) 238
(c) 58
(d) 119

Answer:

Construction: Draw a line PQ parallel to AB which is also parallel to CD
Since, PQ || CD
∴ ∠EFC = ∠FEQ = 37^âˆ˜                 [Alternate angles]
Now, ∠AEQ + ∠FEQ = ∠AEF
⇒ ∠AEQ + 37^âˆ˜ = 95^âˆ˜
⇒ ∠AEQ = 58^âˆ˜
Since, PQ || AB
∴∠EAB + ∠AEQ = 180^âˆ˜                 [Angles on the same side of a transversal line are supplementary]
⇒ ∠EAB + 58^âˆ˜ = 180^âˆ˜
⇒ ∠EAB = 122^âˆ˜
∠EAB + Reflex∠EAB = 360^âˆ˜              [Complete angle]
∴ 122^âˆ˜ + (2x)^âˆ˜ = 360^âˆ˜
⇒ 2x = 238
⇒ x = 119
Hence, the correct answer is option (d).

Page No 14.33:

Question 33:

In Fig. 111, if AB || CO then x =
(a) 154
(b) 139
(c) 144
(d) 164

Answer:

Construction: Draw a line PQ parallel to AB which is also parallel to CD
Since, PQ || AB
∴ ∠AME + ∠QEM = 180^âˆ˜                [Angles on the same side of a transversal line are supplementary]
⇒ 139^âˆ˜ + ∠QEM = 180^âˆ˜
⇒ ∠QEM = 41^âˆ˜
Now, ∠QEM + ∠DEQ = ∠MED
⇒ 41^âˆ˜ + ∠DEQ = 67^âˆ˜
⇒ ∠DEQ = 26^âˆ˜
Now, ∠PED + ∠DEQ = 180^âˆ˜                 [Linear Pair angles]
⇒ ∠PED + 26^âˆ˜ = 180^âˆ˜
⇒ ∠PED = 154^âˆ˜
Since, PQ || AB
∴ x^âˆ˜ = ∠PED                                         [Corresponding angles]
⇒ x^âˆ˜ = 154^âˆ˜
⇒ x = 154
Hence, the correct answer is option (a).

Page No 14.33:

Question 34:

In Fig. 112, if AB || CD, then x =
(a) 32
(b) 42
(c) 52
(d) 31

Answer:

Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠CDP + Reflex∠CDP = 360^âˆ˜              [Complete angle]
∴∠CDP + 249^âˆ˜ = 360^âˆ˜
⇒ ∠CDP = 111^âˆ˜
Since, PQ || AB
∴ ∠BAP = ∠APQ                                         [Alternate angles]
⇒ ∠BAP = 28^âˆ˜
Now, ∠APQ + ∠QPD = ∠APD
⇒ 28^âˆ˜ + ∠QPD = (2x + 13)^âˆ˜
⇒ ∠QPD = (2x + 13)^âˆ˜− 28^âˆ˜
Since, PQ || CD
∴ ∠QPD + ∠CDP = 180^âˆ˜                [Angles on the same side of a transversal line are supplementary]
⇒ (2x + 13)^âˆ˜− 28^âˆ˜ + 111^âˆ˜ = 180^âˆ˜
⇒ 2x + 13− 28 + 111 = 180
⇒ 2x = 84
⇒ x = 42
Hence, the correct answer is option (b).

Page No 14.33:

Question 35:

In Fig. 113 if AC || OF and AB || CE, then
(a) x = 145, y = 223
(c) x = 135, y = 233
(b) x = 223, y = 145
(d) x = 233, y = 135

Answer:

Construction: Produce FD towards D to the point M
∠DCA + Reflex∠DCA = 360^âˆ˜              [Complete angle]
∴∠DCA + (y + 15)^âˆ˜ = 360^âˆ˜
⇒ ∠DCA = 345^âˆ˜ − y^âˆ˜
Now,
∠MDC = ∠EDF = 58^âˆ˜                                    [Vertically Opposite angles]
Since, MF || AC
∴ ∠MDC + ∠QPD = 180^âˆ˜                              [Angles on the same side of a transversal line are supplementary]
⇒ 58^âˆ˜ + 345^âˆ˜ − y^âˆ˜ = 180^âˆ˜
⇒ y = 223
∴ ∠DCA = 345^âˆ˜ − 223^âˆ˜ = 122^âˆ˜
Again, ∠BAC + Reflex∠BAC = 360^âˆ˜              [Complete angle]
∴∠BAC + (2x + 12)^âˆ˜ = 360^âˆ˜
⇒ ∠DCA = 348^âˆ˜ − (2x)^âˆ˜
Since, AB || CD
∴ ∠DCA + ∠DCA = 180^âˆ˜                [Angles on the same side of a transversal line are supplementary]
⇒ 348^âˆ˜ − (2x)^âˆ˜+ 122^âˆ˜ = 180^âˆ˜
⇒ (2x)^âˆ˜= 290^âˆ˜
⇒ x = 145
Hence, the correct answer is option (a).

Page No 14.6:

Question 1:

Write down each pair of adjacent angles shown in Fig.

Answer:

Adjacent angles are the angles that have a common vertex and a common arm.
Following are the adjacent angles in the given figure:

$∠ D O C and ∠ B O C ∠ C O B and ∠ B O A$

Page No 14.6:

Question 2:

In Fig., name all the pairs of adjacent angles.

Answer:

In figure (i), the adjacent angles are:

$∠ EBA and ∠ ABC ∠ ACB and ∠ BCF ∠ BAC and ∠ CAD$

In figure (ii), the adjacent angles are:

$∠$ BAD and $∠$ DAC
$∠$ BDA and $∠$ CDA

Page No 14.6:

Question 3:

In figure, write down: (i) each linear pair (ii) each pair of vertically opposite angles.

Answer:

(i) Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.
$∠$ 1 and $∠$ 3
$∠$ 1 and $∠$ 2
$∠$ 4 and $∠$ 3
$∠$ 4 and $∠$ 2
$∠$ 5 and $∠$ 6
$∠$ 5 and $∠$ 7
$∠$ 6 and $∠$ 8
$∠$ 7 and $∠$ 8

(ii) Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.
$∠$ 1 and $∠$ 4
$∠$ 2 and $∠$ 3
$∠$ 5 and $∠$ 8
$∠$ 6 and $∠$ 7

Page No 14.7:

Question 4:

Are the angles 1 and 2 given in Fig. adjacent angles?

Answer:

No, because they have no common vertex.

Page No 14.7:

Question 5:

Find the complement of each of the following angles:
(i) 35°
(ii) 72°
(iii) 45°
(iv) 85°

Answer:

Two angles are called complementary angles if the sum of those angles is 90°.

Complementary angles of the following angles are:

$(i) 90 ° - 35 ° = 55 ° (ii) 90 ° - 72 ° = 18 ° (iii) 90 ° - 45 ° = 45 ° (iv) 90 ° - 85 ° = 5 °$

Page No 14.7:

Question 6:

Find the supplement of each of the following angles:
(i) 70°
(ii) 120°
(iii) 135°
(iv) 90°

Answer:

Two angles are called supplementary angles if the sum of those angles is 180°.
Supplementary angles of the following angles are:

(i) 180° − 70° = 110°
(ii) 180° − 120° = 60°
(iii) 180° − 135° = 45°
(iv) 180° − 90° = 90°

Page No 14.7:

Question 7:

Identify the complementary and supplementary pairs of angles from the following pairs:
(i) 25°, 65°
(ii) 120°, 60°
(iii) 63°, 27°
(iv) 100°, 80°

Answer:

Since
$(i) 25 ° + 65 ° = 90 °, t h e r e f o r e this is c omplementary pair of a ngle . (i i) 120 ° + 60 ° = 180 °, t herefore t his i s s upplementary pair o f a ngle . (i i i) 63 ° + 27 ° = 90 °, t h e r e f o r e this is c omplementary pair of a ngle . (i v) 100 ° + 80 ° = 180 °, t herefore t his i s s upplementary pair o f a ngle .$

Therefore, (i) and (iii) are the pairs of complementary angles and (ii) and (iv) are the pairs of supplementary angles.

Page No 14.7:

Question 8:

Can two angles be supplementary, if both of them be
(i) obtuse?
(ii) right?
(iii) acute?

Answer:

(i) No, two obtuse angles cannot be supplementary.
(ii) Yes, two right angles can be supplementary. ( $∵ ∠ 90 ° + ∠ 90 ° = ∠ 180 °$ )
(iii) No, two acute angles cannot be supplementary.

Page No 14.7:

Question 9:

Name the four pairs of supplementary angles shown in Fig.

Answer:

Following are the supplementary angles:
$∠$ AOC and $∠$ COB
$∠$ BOC and $∠$ DOB
$∠$ BOD and $∠$ DOA
$∠$ AOC and $∠$ DOA

Page No 14.7:

Question 10:

In Fig., A, B, C are collinear points and ∠DBA = ∠EBA.

(i) Name two linear pairs
(ii) Name two pairs of supplementary angles.

Answer:

(i) Linear pairs:
$∠$ ABD and $∠$ DBC
$∠$ ABE and $∠$ EBC

Because every linear pair forms supplementary angles, these angles are:
$∠$ ABD and $∠$ DBC
$∠$ ABE and $∠$ EBC

Page No 14.7:

Question 11:

If two supplementary angles have equal measure, what is the measure of each angle?

Answer:

Let x and y be two supplementary angles that are equal.
$∠ x = ∠ y$
According to the question,
$∠ x + ∠ y = 180 ° \Rightarrow ∠ x + ∠ x = 180 ° \Rightarrow 2 ∠ x = 180 ° \Rightarrow ∠ x = \frac{180 °}{2} = 90 ° ∴ ∠ x = ∠ y = 90 °$

Page No 14.7:

Question 12:

If the complement of an angle is 28°, then find the supplement of the angle.

Answer:

Let x be the complement of the given angle $28 °$ .
$∴ ∠ x + 28 ° = 90 ° \Rightarrow ∠ x = 90 ° - 28 ° = 62 °$
So, supplement of the angle = $180 ° - 62 ° = 118 °$

Page No 14.7:

Question 13:

In Fig. 19, name each linear pair and each pair of vertically opposite angles:

Answer:

Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.

∠

1 and

∠

∠

2 and

∠

∠

3 and

∠

∠

1 and

∠

∠

5 and

∠

∠

6 and

∠

∠

7 and

∠

∠

8 and

∠

∠

9 and

∠

∠

10 and

∠

∠

11 and

∠

∠

12 and

∠

9

Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.

∠

1 and

∠

∠

4 and

∠

∠

5 and

∠

∠

6 and

∠

∠

9 and

∠

∠

10 and

∠

Page No 14.7:

Question 14:

In Fig., OE is the bisector of ∠BOD. If ∠1 = 70°, find the magnitudes of ∠2, ∠3 and ∠4.

Answer:

Since OE is the bisector of $∠$ BOD,
$∴ ∠ DOE = ∠ EOB ∠ 2 + ∠ 1 + ∠ EOB = 180 ° (Linear Pair) ∠ 2 + 2 ∠ 1 = 180 ° (∠ 1 = ∠ EOB) \Rightarrow ∠ 2 = 180 ° - 2 ∠ 1 = 180 ° - 2 \times 70 ° = 180 ° - 140 ° = 40 °$
$∠ 4 = ∠ 2 = 40 ° (Vertically opposite angles) ∠ 3 = ∠ DOB = ∠ 1 + ∠ EOB = 70 ° + 70 ° = 140 ° [∠ 3 = ∠ DOB (Vertically opposite angles)]$

Page No 14.7:

Question 15:

One of the angles forming a linear pair is a right angle. What can you say about its other angle?

Answer:

One angle of a linear pair is the right angle, i.e., 90°.
∴ The other angle = 180°â€‹- 90° = 90â€‹°

Page No 14.7:

Question 16:

One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?

Answer:

If one of the angles of a linear pair is obtuse, then the other angle should be acute; only then can their sum be 180°.

Page No 14.7:

Question 17:

One of the angles forming a linear pair is an acute angle. What kind of angle is the other?

Answer:

In a linear pair, if one angle is acute, then the other angle should be obtuse. Only then their sum can be 180°.

Page No 14.8:

Question 18:

Can two acute angles form a linear pair?

Answer:

No, two acute angles cannot form a linear pair because their sum is always less than 180°.

Page No 14.8:

Question 19:

If the supplement of an angle is 65°; then find its complement.

Answer:

Let x be the required angle.
Then, we have:
x + 65°= 180°
$\Rightarrow$ x = 180° - 65° = 115°

The complement of angle x cannot be determined.

Page No 14.8:

Question 20:

Find the value of x in each of the following figures.

Answer:

(i)
Since $∠ BOA + ∠ BOC = 180 °$ (Linear pair)
$∴ ∠ x = 180 ° - ∠ BOA = 180 ° - 60 ° = 120 °$

(ii)
$Since ∠ QOP + ∠ QOR = 180 ° (Linear pair) ∴ 2 x + 3 x = 180 ° \Rightarrow 5 x = 180 ° \Rightarrow x = \frac{180 °}{5} = 36 °$

(iii)
$Since ∠ LOP + ∠ PON + ∠ NOM = 180 ° (Linear pair) ∴ ∠ PON = 180 ° - ∠ LOP - ∠ NOM \Rightarrow x = 180 ° - 35 ° - 60 ° \Rightarrow x = 180 ° - 95 ° = 85 °$

(iv)
$Since ∠ COD + ∠ DOE + ∠ EOA + ∠ AOB + ∠ BOC = 360 ° (Sum of all angles at a point) ∴ 83 ° + 92 ° + 75 ° + 47 ° + x = 360 ° \Rightarrow 297 ° + x = 360 ° \Rightarrow x = 360 ° - 297 ° = 63 °$

(v)
$2 x ° + x ° + 2 x ° + 3 x ° = 180 ° \Rightarrow 8 x = 180 \Rightarrow x = \frac{180}{8} = 22.5 °$

(vi)
$3 x ° = 105 ° \Rightarrow x = \frac{105}{3} = 35 °$

Page No 14.8:

Question 21:

In Fig. 22, it being given that ∠1 = 65°, find all other angles.

Answer:

$∠ 1 = ∠ 3$           (Vertically opposite angles)
$∴ ∠ 3 = 65 °$
Since $∠ 1 + ∠ 2 = 180 °$        (Linear pair)
$∴ ∠ 2 = 180 ° - 65 ° = 115 °$
$∠ 2 = ∠ 4$           (Vertically opposite angles)
$∴ ∠ 4 = ∠ 2 = 115 °$ and $∠ 3 = 65 °$

Page No 14.9:

Question 22:

In Fig., OA and OB are opposite rays:

(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?

Answer:

$∠$ AOC + $∠$ BOC = 180° (Linear pair)
$\Rightarrow (2 y + 5) + 3 x = 180 ° \Rightarrow 3 x + 2 y = 175 °$
(i) If x = 25°, then
$3 \times 25 ° + 2 y = 175 ° \Rightarrow 75 ° + 2 y = 175 ° \Rightarrow 2 y = 175 ° - 75 ° = 100 ° \Rightarrow y = \frac{100 °}{2} = 50 °$
(ii) If y = 35°, then
$3 x + 2 \times 35 ° = 175 ° \Rightarrow 3 x + 70 ° = 175 ° \Rightarrow 3 x = 175 ° - 70 ° = 105 ° \Rightarrow x = \frac{105 °}{3} = 35 °$

Page No 14.9:

Question 23:

In Fig., write all pairs of adjacent angles and all the linear pairs.

Answer:

Adjacent angles:

$∠ D O A and ∠ D O C ∠ D O C and ∠ B O C$

$∠ A O D and ∠ D O B ∠ B O C and ∠ A O C$

Linear pairs of angles:

$∠ A O D and ∠ D O B ∠ B O C and ∠ A O C$

Page No 14.9:

Question 24:

In Fig. 25, find ∠x. Further find ∠BOC, ∠COD and ∠AOD.

Answer:

$∠ A O D + ∠ D O C + ∠ C O B = 180 ° (L i n e a r p a i r) (x + 10) ° + x ° + (x + 20) ° = 180 ° 3 x + 30 ° = 180 ° 3 x = 180 ° - 30 ° 3 x = 150 ° x = \frac{150 °}{3} = 50 °$
$∠ B O C = x + 20 ° = 50 ° + 20 ° = 70 ° ∠ C O D = x = 50 ° ∠ A O D = x + 10 ° = 50 ° + 10 ° = 60 °$

Page No 14.9:

Question 25:

How many pairs of adjacent angles are formed when two lines intersect in a point?

Answer:

If two lines intersect at a point, then four adjacent pairs are formed, and those pairs are linear as well.

Page No 14.9:

Question 26:

How many pairs of adjacent angles, in all, can you name in Fig.?

Answer:

There are 10 adjacent pairs in the given figure; they are:
$∠ EOD and ∠ DOC ∠ COD and ∠ BOC ∠ COB and ∠ BOA$
$∠ AOB and ∠ BOD ∠ BOC and ∠ COE ∠ COD and ∠ COA ∠ DOE and ∠ DOB$
$∠ EOD and ∠ DOA ∠ EOC and ∠ AOC ∠ AOB and ∠ BOE$

Page No 14.9:

Question 27:

In Fig., determine the value of x.

Answer:

$∠ AOB + ∠ BOC = 180 ° (Linear pair) \Rightarrow 3 x + 3 x = 180 ° \Rightarrow 6 x = 180 ° \Rightarrow x = \frac{180 °}{6} = 30 °$

Page No 14.9:

Question 28:

In Fig., AOC is a line, find x.

Answer:

$∠ AOB + ∠ BOC = 180 ° (Linear pair) \Rightarrow 70 ° + 2 x = 180 ° \Rightarrow 2 x = 180 ° - 70 ° = 110 ° \Rightarrow x = \frac{110 °}{2} = 55 °$

Page No 14.9:

Question 29:

In Fig., POS is a line, find x.

Answer:

$∠ QOP + ∠ QOR + ∠ ROS = 180 °$ (Angles on a straight line)

$\Rightarrow 60 ° + 4 x + 40 ° = 180 ° \Rightarrow 100 ° + 4 x = 180 ° \Rightarrow 4 x = 180 ° - 100 ° = 80 ° \Rightarrow x = \frac{80 °}{4} = 20 °$

Page No 14.9:

Question 30:

In Fig., lines l₁ and l₂ intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, z and u.

Answer:

$∠ z = ∠ x = 45 ° (Vertically opposite angles) Now, ∠ x + ∠ y = 180 ° (Linear pair) \Rightarrow ∠ y = 180 ° - 45 ° = 135 ° ∠ u = ∠ y = 135 ° (Vertically opposite angles)$

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