0

Exercise 6.3

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

Sol.         ∠PQT  = ∠ QPR  + ∠PRQ     [Exteric-r angle of a triangle is equal to sum of interior opposite angles]

⇒   on 110° = ∠ QPR + ∠PRQ                                                 ……(i)
Also, ∠SPH + ∠QPR = 180°                                               [Linear pain]

⇒         135° + ∠QPR = 180°  ⇒  ∠ QPR 180° – 135° = 45°.
Substituting the value of ∠QPR in (i), we get
110° = 45° + ∠PRQ  ⇒  ∠PRQ = 110° – 45° = 65°.

2. In figure, X = 62°,  ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XYZ respectively of , find  ∠OZY and ∠YOZ.

Sol. In ΔXYZ, ∠X + ∠XYZ + ∠YZX = 180°                               [Angle sum property]

⇒  62° + 54° + ∠ YZX = 180°
⇒  ∠ YZX = 180° – 116° = 64°                                                   …..(i)
Also, ∠YZX = 2∠OZY                                            [ OZ is bisector of ∠YZX]

⇒  2∠OZY = 64°  ⇒  ∠OZY = 32°                                         …(ii) [From (i)]

Also, $\angle OYZ = {1 \over 2}\angle XYZ = {1 \over 2} \times 54^\circ = 27^\circ$   …(iii)

[OY is bisector of ∠XYZ]

In ΔOYZ, ∠OYZ + ∠OZY + ∠YOZ = 180°        [Sum of angles of a triangle is 180°.]

⇒ 27° + 32° +  ∠YOZ = 180°

⇒ ∠YOZ = 180°  –  59° = 121°
Thus, ∠OZY = 32° and ∠YOZ = 121°

3. In figure, if $AB\parallel DE$, ∠BAC = 35° and ∠CDE = 53­°, find ∠DCE.

Sol. $AB\parallel DE$ and AE is transversal.
∴    ∠DEC = ∠BAC = 35° .                                                                               ..(i)
Now, in triangle CDE,
∠ CDE = 53°                                                                             ……(ii) [Given]

and ∠DCE + ∠DEC + ∠CDE = 180°                           [Sum of angles of a triangle is 180″.]

⇒  ∠DCE 35′ + 53° = 180°                                                           [From (i) and (ii)]

⇒  ∠ DCE = 180°  –  88° = 92°.

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

Sol. In ΔPRT, ∠P + ∠R + ∠PTR. = 180°                              [Sum of angles of a triangle is 180°.]

⇒ 95° + 40° + ∠PTR = 180°
⇒  ∠PTR = 180° — 135° = 45°                                                      …(i)

Also,          ∠STQ = ∠PTR                                                         [Vertically opposite angles]

⇒    ∠ STQ = 45°.                                                                    …(ii) [From (i)]
In ΔTSQ, ∠STQ + ∠S + ∠TQS = 180°                                [Sum of angles of a triangle is 180°.]

⇒              45° + 75° + ∠ TQS = 180°
⇒       ∠TQS = 180° — 120° = 60°.

5. In figure, if $PQ \bot PS$ , $PQ\parallel SR$ , ∠ SQR = 28° and ∠ QRT = 65°, then find the values of x and y.

Sol. $PQ\parallel SR$ and QR is transversal.
∴                    ∠ PQR =  ∠QRT                                                               [Alternate angles]

⇒           ∠ PQS +  ∠SQR = ∠QRT
⇒                x + 28° = 65° ⇒ x = 65° — 28° = 37°                        …..(i)

In right-angled triangle SPQ,
∠P + y + x 180°
[Sum of angles of a triangle is 180°.]

⇒    90° + y + 37° = 180°

⇒   y   =   180°    —    127° = 53° .

6.  In figure, the side QR of  ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, thea prove that
$\angle QTR = {1 \over 2}\angle QPR$

Sol.  $\angle PRT = \angle TRS = {1 \over 2}\angle PRS$                ……..(i)
[ TR is bisector of ∠ PRS]
and  $\angle PQT = \angle TQR = {1 \over 2}\angle PQR$         ………..(ii)
[ TQ is bisector of ∠PQR]
Also ,  ∠PRS = ∠QPR + ∠PQR                                                               ………(iii)
[Exterior angle of a triangle is equal to sum of interior opposite angles.]

and ∠TRS = ∠QTR + ∠TQR                                          ….(iv)    [Reason same as above ]

From (i), (iii) and (iv)
∠QPR + ∠PQR = 2(∠QTR + ∠TQR)
⇒  ∠PQR + ∠PQR = 2 ∠QTR + 2 ∠ TQR
⇒  ∠QPR + ∠PQR = 2 ∠QTR + ∠PQR                              [From (ii)]

⇒                                     ∠ QPR  = 2 ∠QTR

⇒             $\angle QTR = {1 \over 2}\angle QPR$