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EXERCISE 4 A ANGLES, LINES AND TRIANGLES rsmaths99.com

Q1. Define the following term :

(i)                 Angles : When two rays OA and OB meet at a point O, then LAOB is called an angle.

(ii)               Interior of an angle : The interior of an angle is a 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)             Obtuse angle : An angle whose measure is more than 90O but less than 180O is called an obtuse angle.rsmaths99.com

(iv)              Reflex angle : An angle whose measure is more than 180O but less than 360O is called reflex angle.

(v)                Complementary angles : Two angles are said to be complementary angles if their sum is 90O.

(vi)              Supplementary angles : Two angles are said to be supplementary angles if their sum is 180O..

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Q2. If LA = 35O 2746 and LB = 28O4339, find LA + LB.

Sol. LA + LB = 35O 2746 + 28O4339 = 64O70 85

Since 60 = 1 and 60 = 1O rsmaths99.com

LA + LB = 65O1125 Ans.

Q3. Find the difference between two angles measuring 36O and 24O2830.

Sol. 36O - 24O2830 = 35O5960 - 24O2830 =

Deg min sec

35O 59 60

- 24O 28 30

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11O 31 30 rsmaths99.com

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Q4. We know that two angles are complementary if their sum is 90O and two complementary angles are called the complement of each other.

i] complement of 58O = 90O 58O = 32O.

iii] of a right angle = x 90O = 60O rsmaths99.com

complement of 60O = 90O 60O = 30O

v] complement of 52O4320 = 90O - 52O4320

= 89O 5960 - 52O4320 = 37O 1640 . rsmaths99.com

 

Q5. We know that two angles are said to be supplement to each other of their sum is 180O

i] Supplement of 68O = 180O 68O = 112O.

iii] Supplement of x 90O = 54O

= 180O 54O = 126O rsmaths99.com

iv] Supplement of 75O36 = 179O60 - 75O36 = 104O24

v] Supplement of 124O2040 = 179O5960 - 124O2040 = 55O3920

vi] Supplement of 108O4832 = 179O5960 - 108O4832 = 71O1128

 

Q6. Let the measure of required angle = y

Its complement = 90O y

According to the condition, y = 90O y => 2y = 90O => y = 45O

Required angle = 45O. rsmaths99.com

ii] Let the measure of required angle = y

Its supplement = 180O y

According to the condition, y = 180O y => 2y = 180O => y = 90O

Required angle = 90O.

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Q7. Let the required angle = y

Its complement = 90O y

According to the condition, y - (90O - Y) = 36O

y - 90O + y = 36O => 2y = 36O = 90O = 126O => Y = 63O

Required angle = 63O rsmaths99.com

 

Q8. Let the required angle = y

Its supplement = 180O y

According to the condition, (180O y) - Y = 25O

180O y y = 25O => -2Y = 25O 180O => -2y = -155O => y = 77.5O

Required angle = 77.5O

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Q9. Let the required angle = y

Its complement = 90O y

According to the condition, y = 4(90O - y) => Y = 360O 4Y

  Y + 4y = 360O => 5y = 360O => Y = 72O

Required angle = 72O Ans. rsmaths99.com

 

Q13. Let one angle = y, then, its supplement = 180O y

According to the condition, y : (180O y) = 3 : 2

  => 2y = 3(180O - y) => 2y = 540O - 3y

=> 2y + 3y = 540O => 5y = 540O => y = 108O rsmaths99.com

Angles = 108O and its supplement = 180O 108O = 72O

 

Q15. Let the required angle = y

Its complement = 90O y and supplement = 180o y

According to the condition, rsmaths99.com

7(90O - y) = 3(180O y) 10O

  630O 7y = 3(180O y) 10O

  630O 7y = 540O 3y 10O

  -7y + 3y = 540O 10O 630O => -4y = -100O => y = 25O

Required angle = 25O.

 

EXERCISE 4 B rsmaths99.com

 

Q1. AOB is a straight line

LAOC + LBOC = 180O (Linear Pair)

  62O + y = 180O => y = 180O 62O = 118O

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Q2. AOB is a st. line

LAOC + LCOD + LDOB = 180O rsmaths99.com

  (3y 5)O + 55O + (y + 20)O = 180O

  3y 5O + 55O + y + 20O = 180O => 4y = 180O - 70O = 110O

  y = 27.5O

And LAOC = 3y 5O = 3 x 27.5O 5O = 82.5O 5O = 77.5O

LBOD = y + 20O = 27.5O + 20O = 47.5O Ans.

Q3. AOB is a st. line rsmaths99.com

LAOC + LCOD + LDOB = 180O

  (3x + 7)O + (2 x - 19)O + x = 180O

  3 x + 7O + 2 x - 19O + x = 180O => 6 x 12O = 180O => x = 32O

x = 32O rsmaths99.com

LAOC = 3x + 7O = 3 x 32O + 7O

= 96O + 7O = 103O

LCOD = 2x 19O = 2 x 32O - 19O

= 64O 19O 45O rsmaths99.com

LDOB = x = 32O Ans.

Q4. Since XOY is a st. line x + y + z = 180O

But x : y : z = 5 : 4 : 6 rsmaths99.com

Let LXOP = xO = 5a , LPOQ = yO = 4a, LQOY = zO = 6a

Then, 5a + 4a + 6a = 180O => 15a = 180O => a = 12O

x = 5a = 5 x 12O = 60O

y = 4a = 4 x 12O = 48O rsmaths99.com

z = 6a = 6 x 12O = 72O Ans.

Q6. AB and CD intersect each other at O

L AOC = LBOD and LBOC = LAOD (vertically opposite angles)

Now, LAOC = 50O rsmaths99.com

LBOD = LAOC = 50O

Since LAOC + LBOC = 180O (Linear pair)

  50O + LBOC = 180O => LBOC = 130O

LAOD = LBOC = 130O rsmaths99.com

  Hence LBOD = 50O, LBOC = 130O, LAOD = 130O

Q7. Here, AB, CD, and EF are coplanar lines intersecting at O.

LAOF = LBOE, LDOF = LCOE and LBOD = LAOC (vert. opp. Angles)

x = y, z = 50O, t = 90O rsmaths99.com

Now, LAOF + LDOF + LBOD = 180O (angles on the same side of a st. line)

  X + 50O + 90O = 180O => x = 180O 140O = 40O

Hence, x = 40O, y = x = 40O , z = 50O , and t = 90O Ans.

Q8. Three coplanar lines AB, CD and EF intersect at point O.

LAOD = LBOC , LDOF = LCOE and LAOE = LBOF (Vertically opp. angles)

Now, LAOD = 2x LBOC = 2x and LBOF = 3x LAOE = 3x ,

and LCOE = 5x LDOF = 5x rsmaths99.com

But LAOD + LDOF + LBOF + LBOC + LCOE + LAOE = 360O (Angles at a point)

  2x + 5x + 3x + 2x + 5x + 3x = 360O

  20x = 360O => x = 18O rsmaths99.com

LAOD = 2x = 2x 18O = 36O , LCOE = 5x = 5 x 18O = 90O and

LAOE = 3x = 3 x 18O = 54O Ans.

Q9. rsmaths99.com

AOB is a line and CO stands on it forming LAOC and LBOC

Now, LAOC : LBOC = 5 : 4 rsmaths99.com

Let LAOC = 5x and LBOC = 4x

But LAOC + LBOC = 180O (Linear pair)

  5x + 4x = 180O => 9x = 180O => x = 2oO

LAOC = 5x = 5 x 20O = 100O

and LBOC = 4x = 4 x 20O = 80O Ans. rsmaths99.com

Q10.

Two lines AB and CD intersect each other at O and LAOC = 90O

Since LAOC = LBOD (Vert. opp. angles)

LBOD = 90O rsmaths99.com

But LAOC + LBOC = 180O (Linear pair)

  90O + LBOC = 180O => LBOC = 90O

But LAOD = LBOC (vert. opp. angles)

LAOD = 90O . Hence each of the remaining angles is 90O .

Q11. Given, two lines AB and CD intersect each other at O and

LBOC + LAOD = 208O rsmaths99.com

Since LAOD = LBOC (vertically opp. Angles)

LBOC + LBOC = 280O (since LAOD = LBOC)

  2 LBOC = 280O => LBOC = 140O

But LBOC + LAOC = 180O (Linear pair)

  140O + LAOC = 180O rsmaths99.com

  LAOC = 180O 1400 = 40O

But LBOD = LAOC (vertically opp. angles)

LBOD = 40O rsmaths99.com

Hence LAOC = 40O , LBOC = 140O , LBOD = 40O

Q12. OC is the bisector of LAOB and OD is the ray opp. To OC.

Now, LAOC = LBOC [Since OC is bisector of LAOB]

But LBOC + LBOD = 180O [Linear pair]

Similarly, LAOD + LAOC = 180O rsmaths99.com

  LBOC + LBOD = LAOD + LAOC

But LAOC = L BOC [Given]

L BOD = L AOD L AOD = L BOD Proved.

Q13. Since AB is the mirror .PQ is the incident ray, QR is its reflected ray.

  L BQR = L PQA rsmaths99.com

But L BQR + L PQR + L PQA = 180O [Angles on one side of a st. line]

  L PQA + LPQA + 112O = 180O

  2 L PQA + 112O = 180O => 2 L PQA = 180O 112O = 64O

L PQA = 34O Ans.

Q14. rsmaths99.com


Given, Two lines AB and CD intersect each other at O.

OE is the bisector of L BOD and EO is produced to F.

To prove : OF bisects L AOC. rsmaths99.com

Proof : Since AB and CD intersect each other at O.

L AOC = L BOD [vertically opp. angles]

Since OE is the bisector of L BOD

L 1 = L 2 rsmaths99.com

But L 1 = L 3 and L 2 = L 4 [vertically opp. angles]

And L 1 = L 2 [ Proved]

L 3 = L 4 rsmaths99.com

Hence, OF is the bisector of L AOC. Proved


Q15. rsmaths99.com

Given, LAOC and LBOC are supplementary angles

OE is the bisector of LBOC and OF is the bisector of LAOC

To Prove : L1 = L2 , L3 = L 4 [OE and of are the bisectors of LBOC and LAOC respectively]

But LAOC + LBOC = 180O [Linear pair]

  L1 + L2 + L3 + L4 = 180O rsmaths99.com

  L1 + L1 + L3 + L3 = 180O

  2 L1 + 2 L3 = 180O => 2(L1 + L3) = 180O

  L1 + L3 = 90O => EOF = 90O Proved

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