6.5    ANGLE SUM PROPERTY OF A TRIANGLE:

*    The sum of the angles of a triangle is 1800

*    If the side of a triangle is produced then the exterior angle so formed is equal to the sum

of the two interior opposite angles.

Exercise 6.3

1. In Fig. 6.39, sides QP  and  RQ  of  Δ PQR  are produced to points  S  and  T  respectively. If
∠ SPR  =  135°  and  ∠ PQT  =  110°,    find  ∠ PRQ.

Given:     ∠ SPR = 135° and ∠ PQT = 110°,

To find :    ∠ PRQ  =  ?

Proof :     ( i ) ∠ TQP + ∠ PQR = 180°
110° + ∠ PQR = 180°                  (  linear pair )
∴   ∠ PQR  =  180° – 110°  =  70°
( ii ) ∠ SPR + ∠ RPQ  =  180°    ( linear pair )
135° + ∠ RPQ  =  180°
∴ ∠ RPQ  =  180° – 135°  =  45°
We know that
∠ PQR + ∠ PRQ + ∠ RPQ  =  180°   ( Angles of a triangle)

70° + ∠ PRQ + 45° = 180°
∴  ∠ PRQ  =  180° – 115°
∴  ∠ PRQ  =  65°

2. In Fig 6.40, ∠ X = 62°, ∠ XYZ = 54°. If YO and ZO are bisectors of ∠ XYZ and ∠ XZY
respectively of ΔXYZ.   Find ∠ OZY and ∠ YOZ.

Given:   ∠ X = 62°,   ∠ XYZ = 54°
YO and ZO are bisectors of   ∠ XYZ   and  ∠ XZY  respectively.
To find:    ∠ OZY = ?     ∠ YOZ = ?

Proof:    ( i ) ∠ XYZ + ∠ YZX + ∠ X = 180°  ( Angles of a triangle)
54° + ∠ YZX  +  62°  = 180°        ⇒        116° + ∠ YZX = 180°
∴ ∠ YZX  = 180° – 116° = 64°

( ii ) OZ is the bisector of ∠ YZX
∴ ∠ XZO  =  ∠ OZY  =   54°
And ∠ YZX  =  ∠ XZO  +  ∠ OZY  = 64°
∴   ∠ OZY   =   32°                   ( ½  of  64° )
Similarly ∠ OYZ  =  27°         ( ½  of 54° )

( iii ) Now consider ΔYOZ
∠ YOZ + ∠ OZY + ∠ OYZ  = 180° ( Angles of a triangle )
∠ YOZ + 27° + 32° = 180°      ⇒   ∠ YOZ + 59°  =  180°
∠ YOZ = 180° – 59° = 121°      ⇒   ∴ ∠ YOZ  = 121°

 

3. In Fig. 6.41 , if  AB ∥ DE,  ∠ BAC  =  35°  and  ∠ CDE  =  53°,  find  ∠ DCE.

Given: AB ∥ DE     ∠ BAC = 35° , ∠ CDE = 53°
To find :   ∠ DCE  = ?

Proof: (i)   ∠ BAE  =  ∠ AED
( Since AB ∥ DE and AE is transversal, so interior opposite angles are equal )
35°  = ∠ AED                  ( ∵ ∠ BAE = ∠ BAC = 35° )
∴ ∠ CED = 35°            ( interior opposite angles)Consider Δ CDE ,    we get
∠ CDE  +  ∠ DEC  +  ∠ ECD  =  180°   ( Angles of a triangle )
53°  +  35°  +  ∠ ECD  =  180°
88°  +  ∠ ECD  =  180°
∴  ∠ ECD  = 180° – 88°  = 92°
∴  ∠ DCE = 92°

4. In Fig 6.42, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°,  ∠ RPT = 95°
and ∠ TSQ =  75°, find   ∠ SQT.

Given:   ∠ PRT = 40°,  ∠ RPT  =  95° ,  ∠ TSQ  =  75°
To find: ∠ SQT = ?

Proof:  ( i )  Consider Δ PRT
∠ PRT + ∠ RPT + ∠ TPR = 180°   ( Angles of a triangle )
40° + ∠ RPT  +  95°  =  180°             ⇒    135° + ∠ RPT  =  180°
∴  ∠ RPT  =  180°  –  135°  =  45°     ⇒ ∴  ∠ RPT  =45°

( ii ) ∠ RTP = ∠ QTS ( vertically opposite angles )
∴ ∠ QTS = 45°

( iii ) Consider Δ STQ  In the  Fig.6.42
∠ STQ  +  ∠ TQS  +  ∠ QST  =  180°      ( Angles of a triangle )
45°  +  ∠ TQS + 75°  =  180°       ⇒      120° + ∠ TQS  =  180°
∴ ∠ TQS  =  180° − 120°
∴ ∠ TQS  =  60°      and       ∠ SQT  =  60°

5. In Fig. 6.43, if PQ ⊥ PS, PQ∥ SR, ∠ SQR = 28° and ∠ QRT = 65°,                                                                                                 then find the values of   x and y.

Given:   PQ ⊥ PS,   ∠ SQR  =  28° ,  PQ ∥  SR,                                                       ∠ QRT  =  65°,   ∠ QPS  =  90°

To find: values of x and y.

Proof:   ( i ) PQ ∥ SR and QS is the transversal
∠ PQS = ∠ QSR          ( interior opposite angles )
∴   x = ∠ QSR     ( As  ∠ PQS = ∠ QSR  )
∴ ∠ QSR  +  ∠ SQR  =  ∠ QRT
( external angle  =  sum of interior opposite angles of Δ QSR )
x + 28° = 65°    ⇒     x = 65° − 28°     ⇒    ∴ x = 37°

( ii ) Consider Δ QPS , we  get,
∠ QPS + ∠ PSQ + ∠ SQP = 180°    ( Angles of a triangle )
90°  +  y  +  37° =  180°    ⇒       127°  +  y  =  180°
y = 180° – 127°    ⇒           y = 53°

6. In Fig. 6.44, the side QR  of  Δ PQR is produced to a point S. If the bisectors of  ∠ PQR and
∠ PRS  meet at point T, then prove that ∠ QTR  =  ½ ∠ QPR.

Given:  QT is bisector of ∠ PQR  and   RT is bisector of ∠ PRS

To prove: ∠ QTR  =  ½ ∠ QPR.

Proof :  Let   ∠ PQT = ∠ TQR  =  x      ( QT is bisector of ∠ PQR )
and     ∠ PRT  =  ∠ TRS  =  y                  (RT is bisector of ∠ PRS )
( i ) Consider Δ PQR   we get,
∠ QPR + 2 ∠ x = 2 ∠ y    ( exterior angle = sum of interior opposite angles )
∴ 2 ∠ x = 2 ∠ y − ∠ QPR                 – – – – – – – – – ( 1 )
( ii ) Consider Δ TQR   we get,
∠ QTR + ∠ x  =  ∠ y    ( exterior angle = sum of interior opposite angles )
2 ∠ QTR + 2 ∠ x = 2 ∠ y               ( multiply by 2 )
∴ 2 ∠ x  =  2 ∠ y − 2 ∠ QTR               – – – – – – – – – – – ( 2 )
From ( 1 ) and ( 2 ) we get
( iii ) 2 ∠ y  − ∠ QPR  = 2 ∠ y  −  2 ∠ QTR
∠ QPR  = − 2 ∠ QTR      ⇒          ∠ QPR  =  2 ∠ QTR
∴ ∠ QTR  =  ½ ∠ QPR.