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Page No 198:
Question 1:
Answer:
(i) Two rays OA and OB, with a common end-point O, form an angle AOB that is represented as .
(ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A.
(iii) An angle greater than but less than is called an obtuse angle.
(iv) An angle greater than but less than is called a reflex angle.
(v) Two angles are said to be complementary if the sum of their measures is .
(vi) Two angles are said to be supplementary if the sum of their measures is .
Page No 198:
Question 2:
(i) Two rays OA and OB, with a common end-point O, form an angle AOB that is represented as .
(ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A.
(iii) An angle greater than but less than is called an obtuse angle.
(iv) An angle greater than but less than is called a reflex angle.
(v) Two angles are said to be complementary if the sum of their measures is .
(vi) Two angles are said to be supplementary if the sum of their measures is .
Answer:
Two angles whose sum is 90° are called complementary angles.
(i) Complement of 55° = 90° − 55° = 35°
(ii) Complement of
(iii) Complement of 90° = 90° − 90° = 0°
(iv)
Page No 198:
Question 3:
Two angles whose sum is 90° are called complementary angles.
(i) Complement of 55° = 90° − 55° = 35°
(ii) Complement of
(iii) Complement of 90° = 90° − 90° = 0°
(iv)
Answer:
Two angles whose sum is 180° are called supplementary angles.
(i) Supplement of 42° = 180° − 42° = 138°
(ii) Supplement of 90° = 180° − 90° = 90°
(iii) Supplement of 124° = 180° − 124° = 56°
(iv)
Supplement of
Page No 198:
Question 4:
Two angles whose sum is 180° are called supplementary angles.
(i) Supplement of 42° = 180° − 42° = 138°
(ii) Supplement of 90° = 180° − 90° = 90°
(iii) Supplement of 124° = 180° − 124° = 56°
(iv)
Supplement of
Answer:
(i) Let the measure of the required angle be .
Then, in case of complementary angles:
Hence, measure of the angle that is equal to its complement is .
(ii) Let the measure of the required angle be â.
Then, in case of supplementary angles:
Hence, measure of the angle that is equal to its supplement is .
Page No 198:
Question 5:
(i) Let the measure of the required angle be .
Then, in case of complementary angles:
Hence, measure of the angle that is equal to its complement is .
(ii) Let the measure of the required angle be â.
Then, in case of supplementary angles:
Hence, measure of the angle that is equal to its supplement is .
Answer:
Let the measure of the required angle be .
Then, measure of its complement .
Therefore,
Hence, the measure of the required angle is .
Page No 198:
Question 6:
Let the measure of the required angle be .
Then, measure of its complement .
Therefore,
Hence, the measure of the required angle is .
Answer:
Let the measure of the angle be x°.
∴ Supplement of x° = 180° − x°
It is given that,
(180° − x°) − x° = 30°
⇒ 180° − 2x°= 30°
⇒ 2x° = 180° − 30° = 150°
⇒ x° = 75°
Thus, the measure of the angle is 75°.
Page No 198:
Question 7:
Let the measure of the angle be x°.
∴ Supplement of x° = 180° − x°
It is given that,
(180° − x°) − x° = 30°
⇒ 180° − 2x°= 30°
⇒ 2x° = 180° − 30° = 150°
⇒ x° = 75°
Thus, the measure of the angle is 75°.
Answer:
Let the measure of the required angle be .
Then, measure of its complement .
Therefore,
Hence, the measure of the required angle is .
Page No 198:
Question 8:
Let the measure of the required angle be .
Then, measure of its complement .
Therefore,
Hence, the measure of the required angle is .
Answer:
Let the measure of the required angle be .
Then, measure of its supplement .
Therefore,
Hence, the measure of the required angle is .
Page No 198:
Question 9:
Let the measure of the required angle be .
Then, measure of its supplement .
Therefore,
Hence, the measure of the required angle is .
Answer:
Let the measure of the required angle be .
Then, measure of its complement .
And, measure of its supplement.
Therefore,
Hence, the measure of the required angle is .
Page No 198:
Question 10:
Let the measure of the required angle be .
Then, measure of its complement .
And, measure of its supplement.
Therefore,
Hence, the measure of the required angle is .
Answer:
Let the measure of the required angle be .
Then, the measure of its complement .
And the measure of its supplement.
Therefore,
Hence, the measure of the required angle is .
Page No 198:
Question 11:
Let the measure of the required angle be .
Then, the measure of its complement .
And the measure of its supplement.
Therefore,
Hence, the measure of the required angle is .
Answer:
Let the two angles be 4x and 5x, respectively.
Then,
Hence, the two angles are .
Page No 198:
Question 12:
Let the two angles be 4x and 5x, respectively.
Then,
Hence, the two angles are .
Answer:
Two angles whose sum is 90° are called complementary angles.
It is given that the angles (2x – 5)° and (x – 10)° are the complementary angles.
∴ (2x – 5)° + (x – 10)° = 90°
⇒ 3x° – 15° = 90°
⇒ 3x° = 90° + 15° = 105°
⇒ x° = = 35°
Thus, the value of x is 35.
Page No 206:
Question 1:
Two angles whose sum is 90° are called complementary angles.
It is given that the angles (2x – 5)° and (x – 10)° are the complementary angles.
∴ (2x – 5)° + (x – 10)° = 90°
⇒ 3x° – 15° = 90°
⇒ 3x° = 90° + 15° = 105°
⇒ x° = = 35°
Thus, the value of x is 35.
Answer:
We know that the sum of angles in a linear pair is .
Therefore,
Hence, the value of x is .
Page No 206:
Question 2:
We know that the sum of angles in a linear pair is .
Therefore,
Hence, the value of x is .
Answer:
As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is .
Therefore,
Hence,
and
Page No 207:
Question 3:
As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is .
Therefore,
Hence,
and
Answer:
AOB is a straight line. Therefore,
Therefore,
Page No 207:
Question 4:
AOB is a straight line. Therefore,
Therefore,
Answer:
Let
XOY is a straight line. Therefore,
Therefore,
Page No 207:
Question 5:
Let
XOY is a straight line. Therefore,
Therefore,
Answer:
AOB will be a straight line if
Hence, x = 28 will make AOB a straight line.
Page No 207:
Question 6:
AOB will be a straight line if
Hence, x = 28 will make AOB a straight line.
Answer:
We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore,
Let
Also, we know that the sum of all angles around a point is .
Therefore,
Hence,
Therefore, .
Page No 207:
Question 7:
We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore,
Let
Also, we know that the sum of all angles around a point is .
Therefore,
Hence,
Therefore, .
Answer:
We know that if two lines intersect, then the vertically opposite angles are equal.
Hence,
Also,
Hence,
Since, AOB is a straight line, we have:
Also,
Hence,
Page No 207:
Question 8:
We know that if two lines intersect, then the vertically opposite angles are equal.
Hence,
Also,
Hence,
Since, AOB is a straight line, we have:
Also,
Hence,
Answer:
We know that if two lines intersect, then the vertically-opposite angles are equal.
Since, AOB is a straight line, we have:
Therefore,
Page No 207:
Question 9:
We know that if two lines intersect, then the vertically-opposite angles are equal.
Since, AOB is a straight line, we have:
Therefore,
Answer:
Let the two adjacent angles be 5x and 4x, respectively.
Then,
Hence, the two angles are .
Page No 207:
Question 10:
Let the two adjacent angles be 5x and 4x, respectively.
Then,
Hence, the two angles are .
Answer:
We know that if two lines intersect, then the vertically-opposite angles are equal.
And let
Also, we know that the sum of all angles around a point is
Hence,
Hence, the measure of each of the remaining angles is 90o.
Page No 208:
Question 11:
We know that if two lines intersect, then the vertically-opposite angles are equal.
And let
Also, we know that the sum of all angles around a point is
Hence,
Hence, the measure of each of the remaining angles is 90o.
Answer:
We know that if two lines intersect, then the vertically-opposite angles are equal.
Let
Then,
Also, let
We know that the sum of all angles around a point is .
Hence,
Page No 208:
Question 12:
We know that if two lines intersect, then the vertically-opposite angles are equal.
Let
Then,
Also, let
We know that the sum of all angles around a point is .
Hence,
Answer:
Let ∠AOC = 5k and ∠AOD = 7k, where k is some constant.
Here, ∠AOC and ∠AOD form a linear pair.
∴ ∠AOC + ∠AOD = 180º
⇒ 5k + 7k = 180º
⇒ 12k = 180º
⇒ k = 15º
∴ ∠AOC = 5k = 5 × 15º = 75º
∠AOD = 7k = 7 × 15º = 105º
Now, ∠BOD = ∠AOC = 75º (Vertically opposite angles)
∠BOC = ∠AOD = 105º (Vertically opposite angles)
Page No 208:
Question 13:
Let ∠AOC = 5k and ∠AOD = 7k, where k is some constant.
Here, ∠AOC and ∠AOD form a linear pair.
∴ ∠AOC + ∠AOD = 180º
⇒ 5k + 7k = 180º
⇒ 12k = 180º
⇒ k = 15º
∴ ∠AOC = 5k = 5 × 15º = 75º
∠AOD = 7k = 7 × 15º = 105º
Now, ∠BOD = ∠AOC = 75º (Vertically opposite angles)
∠BOC = ∠AOD = 105º (Vertically opposite angles)
Answer:
In the given figure,
∠AOC = ∠BOD = 40º (Vertically opposite angles)
∠BOF = ∠AOE = 35º (Vertically opposite angles)
Now, ∠EOC and ∠COF form a linear pair.
∴ ∠EOC + ∠COF = 180º
⇒ (∠AOE + ∠AOC) + ∠COF = 180º
⇒ 35º + 40º + ∠COF = 180º
⇒ 75º + ∠COF = 180º
⇒ ∠COF = 180º − 75º = 105º
Also, ∠DOE = ∠COF = 105º (Vertically opposite angles)
Page No 208:
Question 14:
In the given figure,
∠AOC = ∠BOD = 40º (Vertically opposite angles)
∠BOF = ∠AOE = 35º (Vertically opposite angles)
Now, ∠EOC and ∠COF form a linear pair.
∴ ∠EOC + ∠COF = 180º
⇒ (∠AOE + ∠AOC) + ∠COF = 180º
⇒ 35º + 40º + ∠COF = 180º
⇒ 75º + ∠COF = 180º
⇒ ∠COF = 180º − 75º = 105º
Also, ∠DOE = ∠COF = 105º (Vertically opposite angles)
Answer:
Here, ∠AOC and ∠BOC form a linear pair.
∴ ∠AOC + ∠BOC = 180º
⇒ xº + 125º = 180º
⇒ xº = 180º − 125ºâ = 55ºâ
Now,
∠AOD = ∠BOC = 125º (Vertically opposite angles)
∴ yº = 125ºâ
∠BOD = ∠AOC = 55º (Vertically opposite angles)
∴ zº = 55ºâ
Thus, the respective values of x, y and z are 55, 125 and 55.
Page No 208:
Question 15:
Here, ∠AOC and ∠BOC form a linear pair.
∴ ∠AOC + ∠BOC = 180º
⇒ xº + 125º = 180º
⇒ xº = 180º − 125ºâ = 55ºâ
Now,
∠AOD = ∠BOC = 125º (Vertically opposite angles)
∴ yº = 125ºâ
∠BOD = ∠AOC = 55º (Vertically opposite angles)
∴ zº = 55ºâ
Thus, the respective values of x, y and z are 55, 125 and 55.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Let .
We know that vertically-opposite angles are equal.
But, [Since OE bisects ]
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.
Page No 208:
Question 16:
![](/img/site_content/ask-answer/loader.gif)
Let .
We know that vertically-opposite angles are equal.
But, [Since OE bisects ]
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Then, we have:
Let .
Then, we have:
Therefore,
Page No 223:
Question 1:
![](/img/site_content/ask-answer/loader.gif)
Then, we have:
Let .
Then, we have:
Therefore,
Answer:
We have, . Then,
Page No 223:
Question 2:
We have, . Then,
Answer:
In the given figure, ∠7 and ∠8 form a linear pair.
∴ ∠7 + ∠8 = 180º
⇒ 80º + ∠8 = 180º
⇒ ∠8 = 180º − 80º = 100º
Now,
∠6 = ∠8 = 100º (Vertically opposite angles)
∠5 = ∠7 = 80º (Vertically opposite angles)
It is given that, l || m and t is a transversal.
∴ ∠1 = ∠5 = 80º (Pair of corresponding angles)
∠2 = ∠6 = 100º (Pair of corresponding angles)
∠3 = ∠7 = 80º (Pair of corresponding angles)
∠4 = ∠8 = 100º (Pair of corresponding angles)
Page No 223:
Question 3:
In the given figure, ∠7 and ∠8 form a linear pair.
∴ ∠7 + ∠8 = 180º
⇒ 80º + ∠8 = 180º
⇒ ∠8 = 180º − 80º = 100º
Now,
∠6 = ∠8 = 100º (Vertically opposite angles)
∠5 = ∠7 = 80º (Vertically opposite angles)
It is given that, l || m and t is a transversal.
∴ ∠1 = ∠5 = 80º (Pair of corresponding angles)
∠2 = ∠6 = 100º (Pair of corresponding angles)
∠3 = ∠7 = 80º (Pair of corresponding angles)
∠4 = ∠8 = 100º (Pair of corresponding angles)
Answer:
Let ∠1 = 2k and ∠2 = 3k, where k is some constant.
Now, ∠1 and ∠2 form a linear pair.
∴ ∠1 + ∠2 = 180º
⇒ 2k + 3k = 180º
⇒ 5k = 180º
⇒ k = 36º
∴ ∠1 = 2k = 2 × 36º = 72º
∠2 = 3k = 3 × 36º = 108º
Now,
∠3 = ∠1 = 72º (Vertically opposite angles)
∠4 = ∠2 = 108º (Vertically opposite angles)
It is given that, l || m and t is a transversal.
∴ ∠5 = ∠1 = 72º (Pair of corresponding angles)
∠6 = ∠2 = 108º (Pair of corresponding angles)
∠7 = ∠1 = 72º (Pair of alternate exterior angles)
∠8 = ∠2 = 108º (Pair of alternate exterior angles)
Page No 223:
Question 4:
Let ∠1 = 2k and ∠2 = 3k, where k is some constant.
Now, ∠1 and ∠2 form a linear pair.
∴ ∠1 + ∠2 = 180º
⇒ 2k + 3k = 180º
⇒ 5k = 180º
⇒ k = 36º
∴ ∠1 = 2k = 2 × 36º = 72º
∠2 = 3k = 3 × 36º = 108º
Now,
∠3 = ∠1 = 72º (Vertically opposite angles)
∠4 = ∠2 = 108º (Vertically opposite angles)
It is given that, l || m and t is a transversal.
∴ ∠5 = ∠1 = 72º (Pair of corresponding angles)
∠6 = ∠2 = 108º (Pair of corresponding angles)
∠7 = ∠1 = 72º (Pair of alternate exterior angles)
∠8 = ∠2 = 108º (Pair of alternate exterior angles)
Answer:
For the lines l and m to be parallel
Page No 224:
Question 5:
For the lines l and m to be parallel
Answer:
Page No 224:
Question 6:
Answer:
and CD is the transversal.
Then,
and BC is the transversal.
Page No 224:
Question 7:
and CD is the transversal.
Then,
and BC is the transversal.
Answer:
and CE is the transversal.
Then,
Again, and BC is the transversal.
Then,
Page No 224:
Question 8:
and CE is the transversal.
Then,
Again, and BC is the transversal.
Then,
Answer:
![](/img/site_content/ask-answer/loader.gif)
and let EF and EG be the transversals.
Now, and EF is the transversal.
Then,
Also,
And,
We know that the sum of angles of a triangle is
Page No 224:
Question 9:
![](/img/site_content/ask-answer/loader.gif)
and let EF and EG be the transversals.
Now, and EF is the transversal.
Then,
Also,
And,
We know that the sum of angles of a triangle is
Answer:
(i)
Draw .
Now, and BE is the transversal.
Then,
Again, and DE is the transversal.
Then,
(ii)
Draw .
Then,
Now, and BO is the transversal.
Again, and DO is the transversal.
Therefore,
(iii)
Draw .
Then,
Now, and AE is the transversal.
Again, and CE is the transversal.
Therefore,
Page No 225:
Question 10:
(i)
Draw .
Now, and BE is the transversal.
Then,
Again, and DE is the transversal.
Then,
(ii)
Draw .
Then,
Now, and BO is the transversal.
Again, and DO is the transversal.
Therefore,
(iii)
Draw .
Then,
Now, and AE is the transversal.
Again, and CE is the transversal.
Therefore,
Answer:
![](/img/site_content/ask-answer/loader.gif)
Draw .
and CE is the transversal.
Then,
Again, and AE is the transversal.
Then,
Page No 225:
Question 11:
![](/img/site_content/ask-answer/loader.gif)
Draw .
and CE is the transversal.
Then,
Again, and AE is the transversal.
Then,
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given, .
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
We know that the sum of angles of a triangle is .
And
Page No 225:
Question 12:
![](/img/site_content/ask-answer/loader.gif)
Given, .
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
We know that the sum of angles of a triangle is .
And
Answer:
and AC is the transversal.
Then,
And,
We know that the sum of the angles of a triangle is .
Page No 225:
Question 13:
and AC is the transversal.
Then,
And,
We know that the sum of the angles of a triangle is .
Answer:
and PQ is the transversal.
Then,
And,
Also,
We know that the sum of angles of a triangle is .
Page No 226:
Question 14:
and PQ is the transversal.
Then,
And,
Also,
We know that the sum of angles of a triangle is .
Answer:
We know that the sum of the angles of a triangle is .
Page No 226:
Question 15:
We know that the sum of the angles of a triangle is .
Answer:
![](/img/site_content/ask-answer/loader.gif)
Draw through E.
Now, and AE is the transversal.
Then,
Again, and CE is the transversal.
Then,
Page No 226:
Question 16:
![](/img/site_content/ask-answer/loader.gif)
Draw through E.
Now, and AE is the transversal.
Then,
Again, and CE is the transversal.
Then,
Answer:
![](/img/site_content/ask-answer/loader.gif)
Draw .
Now, and EF is the transversal.
Then,
Also, .
Page No 226:
Question 17:
![](/img/site_content/ask-answer/loader.gif)
Draw .
Now, and EF is the transversal.
Then,
Also, .
Answer:
In the given figure,
Page No 226:
Question 18:
In the given figure,
Answer:
It is given that, AB || CD and t is a transversal.
∴ ∠BEF + ∠EFD = 180° .....(1) (Sum of the interior angles on the same side of a transversal is supplementary)
EG is the bisector of ∠BEF. (Given)
∴ ∠BEG = ∠GEF = ∠BEF
⇒ ∠BEF = 2∠GEF .....(2)
Also, FG is the bisector of ∠EFD. (Given)
∴ ∠EFG = ∠GFD = ∠EFD
⇒ ∠EFD = 2∠EFG .....(3)
From (1), (2) and (3), we have
2∠GEF + 2∠EFG = 180°
⇒ 2(∠GEF + ∠EFG) = 180°
⇒ ∠GEF + ∠EFG = 90° .....(4)
In âEFG,
∠GEF + ∠EFG + ∠EGF = 180° (Angle sum property)
⇒ 90° + ∠EGF = 180° [Using (4)]
⇒ ∠EGF = 180° − 90° = 90°
Page No 227:
Question 19:
It is given that, AB || CD and t is a transversal.
∴ ∠BEF + ∠EFD = 180° .....(1) (Sum of the interior angles on the same side of a transversal is supplementary)
EG is the bisector of ∠BEF. (Given)
∴ ∠BEG = ∠GEF = ∠BEF
⇒ ∠BEF = 2∠GEF .....(2)
Also, FG is the bisector of ∠EFD. (Given)
∴ ∠EFG = ∠GFD = ∠EFD
⇒ ∠EFD = 2∠EFG .....(3)
From (1), (2) and (3), we have
2∠GEF + 2∠EFG = 180°
⇒ 2(∠GEF + ∠EFG) = 180°
⇒ ∠GEF + ∠EFG = 90° .....(4)
In âEFG,
∠GEF + ∠EFG + ∠EGF = 180° (Angle sum property)
⇒ 90° + ∠EGF = 180° [Using (4)]
⇒ ∠EGF = 180° − 90° = 90°
Answer:
It is given that, AB || CD and t is a transversal.
∴ ∠AEF = ∠EFD .....(1) (Pair of alternate interior angles)
EP is the bisectors of ∠AEF. (Given)
∴ ∠AEP = ∠FEP = ∠AEF
⇒ ∠AEF = 2∠FEP .....(2)
Also, FQ is the bisectors of ∠EFD.
∴ ∠EFQ = ∠QFD = ∠EFD
⇒ ∠EFD = 2∠EFQ .....(3)
From (1), (2) and (3), we have
2∠FEP = 2∠EFQ
⇒ ∠FEP = ∠EFQ
Thus, the lines EP and FQ are intersected by a transversal EF such that the pair of alternate interior angles formed are equal.
∴ EP || FQ (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)
Page No 227:
Question 20:
It is given that, AB || CD and t is a transversal.
∴ ∠AEF = ∠EFD .....(1) (Pair of alternate interior angles)
EP is the bisectors of ∠AEF. (Given)
∴ ∠AEP = ∠FEP = ∠AEF
⇒ ∠AEF = 2∠FEP .....(2)
Also, FQ is the bisectors of ∠EFD.
∴ ∠EFQ = ∠QFD = ∠EFD
⇒ ∠EFD = 2∠EFQ .....(3)
From (1), (2) and (3), we have
2∠FEP = 2∠EFQ
⇒ ∠FEP = ∠EFQ
Thus, the lines EP and FQ are intersected by a transversal EF such that the pair of alternate interior angles formed are equal.
∴ EP || FQ (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)
Answer:
It is given that, BA || ED and BC || EF.
Construction: Extend DE such that it intersects BC at J. Also, extend FE such that it intersects AB at H.
![](/img/site_content/ask-answer/loader.gif)
Now, BA || JD and BC is a transversal.
∴ ∠ABC = ∠DJC .....(1) (Pair of corresponding angles)
Also, BC || HF and DJ is a transversal.
∴ ∠DJC = ∠DEF .....(2) (Pair of corresponding angles)
From (1) and (2), we have
∠ABC = ∠DEF
Page No 227:
Question 21:
It is given that, BA || ED and BC || EF.
Construction: Extend DE such that it intersects BC at J. Also, extend FE such that it intersects AB at H.
![](/img/site_content/ask-answer/loader.gif)
Now, BA || JD and BC is a transversal.
∴ ∠ABC = ∠DJC .....(1) (Pair of corresponding angles)
Also, BC || HF and DJ is a transversal.
∴ ∠DJC = ∠DEF .....(2) (Pair of corresponding angles)
From (1) and (2), we have
∠ABC = ∠DEF
Answer:
It is given that, BA || ED and BC || EF.
Construction: Extend ED such that it intersects BC at G.
![](/img/site_content/ask-answer/loader.gif)
Now, BA || GE and BC is a transversal.
∴ ∠ABC = ∠EGC .....(1) (Pair of corresponding angles)
Also, BC || EF and EG is a transversal.
∴ ∠EGC + ∠GEF = 180° .....(2) (Interior angles on the same side of the transversal are supplementary)
From (1) and (2), we have
â∠ABC + ∠GEF = 180°
Or ∠ABC + ∠DEF = 180°
Page No 227:
Question 22:
It is given that, BA || ED and BC || EF.
Construction: Extend ED such that it intersects BC at G.
![](/img/site_content/ask-answer/loader.gif)
Now, BA || GE and BC is a transversal.
∴ ∠ABC = ∠EGC .....(1) (Pair of corresponding angles)
Also, BC || EF and EG is a transversal.
∴ ∠EGC + ∠GEF = 180° .....(2) (Interior angles on the same side of the transversal are supplementary)
From (1) and (2), we have
â∠ABC + ∠GEF = 180°
Or ∠ABC + ∠DEF = 180°
Answer:
AP is normal to the plane mirror OA and BP is normal to the plane mirror OB.
It is given that the two plane mirrors are perpendicular to each other.
Therefore, BP || OA and AP || OB.
So, BP ⊥ AP (OA ⊥ OB)
⇒ ∠APB = 90° .....(1)
In âAPB,
â∠2 + ∠3 + ∠APB = 180° (Angle sum property)
∴ ∠2 + ∠3 + 90° = 180° [Using (1)]
⇒ ∠2 + ∠3 = 180° − 90° = 90°
⇒ 2∠2 + 2∠3 = 2 × 90° = 180° .....(2)
By law of reflection, we have
∠1 = ∠2 and ∠3 = ∠4 .....(3) (Angle of incidence = Angle of reflection)
From (2) and (3), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠BAC + ∠ABD = 180° (∠1 + ∠2 = ∠BAC and ∠3 + ∠4 = ∠ABD)
Thus, the lines CA and BD are intersected by a transversal AB such that the interior angles on the same side of the transversal are supplementary.
∴ CA || BD
Page No 227:
Question 23:
AP is normal to the plane mirror OA and BP is normal to the plane mirror OB.
It is given that the two plane mirrors are perpendicular to each other.
Therefore, BP || OA and AP || OB.
So, BP ⊥ AP (OA ⊥ OB)
⇒ ∠APB = 90° .....(1)
In âAPB,
â∠2 + ∠3 + ∠APB = 180° (Angle sum property)
∴ ∠2 + ∠3 + 90° = 180° [Using (1)]
⇒ ∠2 + ∠3 = 180° − 90° = 90°
⇒ 2∠2 + 2∠3 = 2 × 90° = 180° .....(2)
By law of reflection, we have
∠1 = ∠2 and ∠3 = ∠4 .....(3) (Angle of incidence = Angle of reflection)
From (2) and (3), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠BAC + ∠ABD = 180° (∠1 + ∠2 = ∠BAC and ∠3 + ∠4 = ∠ABD)
Thus, the lines CA and BD are intersected by a transversal AB such that the interior angles on the same side of the transversal are supplementary.
∴ CA || BD
Answer:
Here, ∠BAC = ∠ACD = 110°
Thus, lines AB aand CD are intersected by a transversal AC such that the pair of alternate angles are equal.
∴ AB || CD (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)
Thus, line AB is parallel to line CD.
Also, ∠ACD + ∠CDE = 110° + 80° = 190° ≠ 180°
If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal are supplementary, then the two lines are parallel.
Therefore, line AC is not parallel to line DE.
Page No 228:
Question 24:
Here, ∠BAC = ∠ACD = 110°
Thus, lines AB aand CD are intersected by a transversal AC such that the pair of alternate angles are equal.
∴ AB || CD (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)
Thus, line AB is parallel to line CD.
Also, ∠ACD + ∠CDE = 110° + 80° = 190° ≠ 180°
If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal are supplementary, then the two lines are parallel.
Therefore, line AC is not parallel to line DE.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Let the two lines m and n be respectively perpendicular to the two parallel lines p and q.
To prove: m is parallel to n.
Proof: Since, m is perpendicular to p
Also, ân is perpendicular to q
Since p and q are parallel and m is a transversal line
[Corresponding angles]
Also,
We know that if two corresponding angles are equal then the two lines containing them must be parallel.
Therefore, the lines m and n are parallel to each other.
Page No 231:
Question 1:
![](/img/site_content/ask-answer/loader.gif)
Let the two lines m and n be respectively perpendicular to the two parallel lines p and q.
To prove: m is parallel to n.
Proof: Since, m is perpendicular to p
Also, ân is perpendicular to q
Since p and q are parallel and m is a transversal line
[Corresponding angles]
Also,
We know that if two corresponding angles are equal then the two lines containing them must be parallel.
Therefore, the lines m and n are parallel to each other.
Answer:
Let âABC be such that ∠A = ∠B + ∠C.
In âABC,
∠A + ∠B + ∠C = 180º (Angle sum property)
⇒ ∠A + ∠A = 180º (∠A = ∠B + ∠C)
⇒ 2∠A = 180º
⇒ ∠A = 90º
Therefore, âABC is a right triangle.
Thus, if one angle of a triangle is equal to the sum of the other two angles, then the triangle is a right triangle.
Hence, the correct answer is option (d).
Page No 232:
Question 2:
Let âABC be such that ∠A = ∠B + ∠C.
In âABC,
∠A + ∠B + ∠C = 180º (Angle sum property)
⇒ ∠A + ∠A = 180º (∠A = ∠B + ∠C)
⇒ 2∠A = 180º
⇒ ∠A = 90º
Therefore, âABC is a right triangle.
Thus, if one angle of a triangle is equal to the sum of the other two angles, then the triangle is a right triangle.
Hence, the correct answer is option (d).
Answer:
Let the measure of each of the two equal interior opposite angles of the triangle be x.
In a triangle, the exterior angle is equal to the sum of the two interior opposite angles.
∴ x + x = 110°
⇒ 2x = 110°
⇒ x = 55°
Thus, the measure of each of these equal angles is 55°.
Hence, the correct answer is option (b).
Page No 232:
Question 3:
Let the measure of each of the two equal interior opposite angles of the triangle be x.
In a triangle, the exterior angle is equal to the sum of the two interior opposite angles.
∴ x + x = 110°
⇒ 2x = 110°
⇒ x = 55°
Thus, the measure of each of these equal angles is 55°.
Hence, the correct answer is option (b).
Answer:
(a) acute-angled
Let the angles measure .
Then,
Therefore, the angles are .
Hence, the triangle is acute-angled.
Page No 232:
Question 4:
(a) acute-angled
Let the angles measure .
Then,
Therefore, the angles are .
Hence, the triangle is acute-angled.
Answer:
Let âABC be such that ∠A = 130°.
![](/img/site_content/ask-answer/loader.gif)
Here, BP is the bisector of ∠B and CP is the bisector of ∠C.
∴ ∠ABP = ∠PBC = ∠B .....(1)
Also, ∠ACP = ∠PCB = ∠C .....(2)
In âABC,
∠A + ∠B + ∠C = 180° (Angle sum property)
⇒ 130° + ∠B + ∠C = 180°
⇒ ∠B + ∠C = 180° − 130° = 50°
⇒ ∠B + ∠C = × 50° = 25°
⇒ ∠PBC + ∠PCB = 25° .....(3) [Using (1) and (2)]
In âPBC,
∠PBC + ∠PCB + ∠BPC = 180° (Angle sum property)
⇒ 25° + ∠BPC = 180° [Using (3)]
⇒ ∠BPC = 180° − 25° = 155°
Thus, if one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles is 155°.
Hence, the correct answer is option (d).
Page No 232:
Question 5:
Let âABC be such that ∠A = 130°.
![](/img/site_content/ask-answer/loader.gif)
Here, BP is the bisector of ∠B and CP is the bisector of ∠C.
∴ ∠ABP = ∠PBC = ∠B .....(1)
Also, ∠ACP = ∠PCB = ∠C .....(2)
In âABC,
∠A + ∠B + ∠C = 180° (Angle sum property)
⇒ 130° + ∠B + ∠C = 180°
⇒ ∠B + ∠C = 180° − 130° = 50°
⇒ ∠B + ∠C = × 50° = 25°
⇒ ∠PBC + ∠PCB = 25° .....(3) [Using (1) and (2)]
In âPBC,
∠PBC + ∠PCB + ∠BPC = 180° (Angle sum property)
⇒ 25° + ∠BPC = 180° [Using (3)]
⇒ ∠BPC = 180° − 25° = 155°
Thus, if one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles is 155°.
Hence, the correct answer is option (d).
Answer:
It is given that, AOB is a straight line.
∴ 60º + (5xº + 3xº) = 180º (Linear pair)
⇒ 8xº = 180º − 60º = 120º
⇒ xº = 15º
Thus, the value of x is 15.
Hence, the correct answer is option (b).
Page No 232:
Question 6:
It is given that, AOB is a straight line.
∴ 60º + (5xº + 3xº) = 180º (Linear pair)
⇒ 8xº = 180º − 60º = 120º
⇒ xº = 15º
Thus, the value of x is 15.
Hence, the correct answer is option (b).
Answer:
Suppose âABC be such that ∠A : ∠B : ∠C = 2 : 3 : 4.
Let ∠A = 2k, ∠B = 3k and ∠C = 4k, where k is some constant.
In âABC,
∠A + ∠B + ∠C = 180º (Angle sum property)
⇒ 2k + 3k + 4k = 180º
⇒ 9k = 180º
⇒ k = 20º
∴ Measure of the largest angle = 4k = 4 × 20º = 80º
Hence, the correct answer is option (c).
Page No 232:
Question 7:
Suppose âABC be such that ∠A : ∠B : ∠C = 2 : 3 : 4.
Let ∠A = 2k, ∠B = 3k and ∠C = 4k, where k is some constant.
In âABC,
∠A + ∠B + ∠C = 180º (Angle sum property)
⇒ 2k + 3k + 4k = 180º
⇒ 9k = 180º
⇒ k = 20º
∴ Measure of the largest angle = 4k = 4 × 20º = 80º
Hence, the correct answer is option (c).
Answer:
In the given figure, OA || CD.
Construction: Extend OA such that it intersects BC at E.
![](/img/site_content/ask-answer/loader.gif)
Now, OE || CD and BC is a transversal.
∴ ∠AEC = ∠BCD = 130° (Pair of corresponding angles)
Also, ∠OAB + ∠BAE = 180° (Linear pair)
∴ 110° + ∠BAE = 180°
⇒ ∠BAE = 180° − 110° = 70°
In âABE,
∠AEC = ∠BAE + ∠ABE (In a triangle, exterior angle is equal to the sum of two opposite interior angles)
∴ 130° = 70° + x°
⇒ x° = 130° − 70° = 60°
Thus, the measure of angle ∠ABC is 60°.
Hence, the correct answer is option (c).
Page No 232:
Question 8:
In the given figure, OA || CD.
Construction: Extend OA such that it intersects BC at E.
![](/img/site_content/ask-answer/loader.gif)
Now, OE || CD and BC is a transversal.
∴ ∠AEC = ∠BCD = 130° (Pair of corresponding angles)
Also, ∠OAB + ∠BAE = 180° (Linear pair)
∴ 110° + ∠BAE = 180°
⇒ ∠BAE = 180° − 110° = 70°
In âABE,
∠AEC = ∠BAE + ∠ABE (In a triangle, exterior angle is equal to the sum of two opposite interior angles)
∴ 130° = 70° + x°
⇒ x° = 130° − 70° = 60°
Thus, the measure of angle ∠ABC is 60°.
Hence, the correct answer is option (c).
Answer:
(a) an acute angle
If two angles are complements of each other, that is, the sum of their measures is , then each angle is an acute angle.
Page No 232:
Question 9:
(a) an acute angle
If two angles are complements of each other, that is, the sum of their measures is , then each angle is an acute angle.
Answer:
An angle which measures more than 180° but less than 360° is called a reflex angle.
Hence, the correct answer is option (d).
Page No 232:
Question 10:
An angle which measures more than 180° but less than 360° is called a reflex angle.
Hence, the correct answer is option (d).
Answer:
(d) 75°
Let the measure of the required angle be .
Then, the measure of its complement will be .
Page No 233:
Question 11:
(d) 75°
Let the measure of the required angle be .
Then, the measure of its complement will be .
Answer:
(b) 54°
Let the measure of the required angle be .
Then, the measure of its complement will be.
Page No 233:
Question 12:
(b) 54°
Let the measure of the required angle be .
Then, the measure of its complement will be.
Answer:
(c) 80°
We have :
Page No 233:
Question 13:
(c) 80°
We have :
Answer:
(b) 86°
We have :
Page No 233:
Question 14:
(b) 86°
We have :
Answer:
(c) 80°
We have :
Page No 233:
Question 15:
(c) 80°
We have :
Answer:
(a) Through a given point, only one straight line can be drawn.
Clearly, statement (a) is false because we can draw infinitely many straight lines through a given point.
Page No 233:
Question 16:
(a) Through a given point, only one straight line can be drawn.
Clearly, statement (a) is false because we can draw infinitely many straight lines through a given point.
Answer:
(b) 30°
Let the measure of the required angle be
Then, the measure of its supplement will be
Page No 233:
Question 17:
(b) 30°
Let the measure of the required angle be
Then, the measure of its supplement will be
Answer:
(a) 60°
Let
Then, we have:
Page No 233:
Question 18:
(a) 60°
Let
Then, we have:
Answer:
(c) 45°
We have :
Page No 234:
Question 19:
(c) 45°
We have :
Answer:
(b) 115°
We have :
Now,
Page No 234:
Question 20:
(b) 115°
We have :
Now,
Answer:
(c) 36°
We know that angle of incidence = angle of reflection.
Then, let
Now,
Page No 234:
Question 21:
(c) 36°
We know that angle of incidence = angle of reflection.
Then, let
Now,
Answer:
(c) 50°
![](/img/site_content/ask-answer/loader.gif)
Draw .
Now, is the transversal.
Also,
is the transversal.
Now,
Page No 234:
Question 22:
(c) 50°
![](/img/site_content/ask-answer/loader.gif)
Draw .
Now, is the transversal.
Also,
is the transversal.
Now,
Answer:
(a) 130°
![](/img/site_content/ask-answer/loader.gif)
Draw
Now, is the transversal.
Also,
is the transversal.
Page No 234:
Question 23:
(a) 130°
![](/img/site_content/ask-answer/loader.gif)
Draw
Now, is the transversal.
Also,
is the transversal.
Answer:
(c) 45°
is the transversal.
Side EC of triangle EFC is produced to D.
Page No 234:
Question 24:
(c) 45°
is the transversal.
Side EC of triangle EFC is produced to D.
Answer:
(b) 126°
Let
Let the transversal intersect AB at P, CD at O and EF at Q.
![](/img/site_content/ask-answer/loader.gif)
Then, we have:
Also,
Page No 235:
Question 25:
(b) 126°
Let
Let the transversal intersect AB at P, CD at O and EF at Q.
![](/img/site_content/ask-answer/loader.gif)
Then, we have:
Also,
Answer:
(a) 50°
is the transversal.
Side QR of traingle PQR is produced to D.
Page No 235:
Question 26:
(a) 50°
is the transversal.
Side QR of traingle PQR is produced to D.
Answer:
(c) 70°
is the transversal.
In , we have:
Page No 235:
Question 27:
(c) 70°
is the transversal.
In , we have:
Answer:
(c) 30°
In , we have:
In , we have:
Page No 235:
Question 28:
(c) 30°
In , we have:
In , we have:
Answer:
(b) 54°
We have:
Also,
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