- In Fig. 6.39, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠ SPR = 135° and ∠ PQT = 110°, find ∠ PRQ.
Solution:
∠RQT = ∠PQT + ∠PQR [Linear pair axiom]
∠PQR = 180° – ∠PQT
= 180° – 110°
∠PQR = 70°
∠SPQ = ∠SPR + ∠RPQ [Linear pair axiom]
∠RPQ = 180° – ∠SPR
= 180° – 135°
∠RPQ = 45°
∠PQR + ∠RPQ + ∠PRQ = 180° [Sum of interior angles of a triangle is 180°]
∠PRQ = 180° – 45° – 70°
∠PRQ = 65°
- In Fig. 6.40, ∠ X = 62°, ∠ XYZ = 54°. If YO and ZO are the bisectors of ∠ XYZ and ∠ XZY respectively of ∆ XYZ, find ∠ OZY and ∠ YOZ.
Solution:
Given ∠x = 62° and ∠XYZ = 54°. If YO and ZO are the bisectors of ∠ XYZ and ∠ XZY respectively of ∆ XYZ. We have to find ∠OZY and ∠YOZ
In ∆ XYZ, ∠XYZ +∠XZY + ∠YXZ = 180° [Sum of interior angles = 180°]
62° + 54° + ∠XZY = 180°
∠XZY = 180° – 62° – 54° = 64°
Since ZO is the bisectors of ∠XZY in ∆ XYZ, then ∠ OZY = ∠ XZO
∠XZY = ∠OZY + ∠XZO
= ∠OZY + ∠OZY
= 2 ∠OZY
64° = 2 ∠OZY
∠OZY = 64°/2 = 32°
Since YO is the bisectors of ∠ XYZ in ∆ XYZ, then ∠ XYO = ∠ OYZ
∠ XYZ = ∠ XYO + ∠ OYZ
= ∠ OYZ + ∠ OYZ
= 2 ∠OYZ
54° = 2 ∠OYZ
∠OYZ = 54°/2 = 27°
∠OZY + ∠OYZ + ∠YOZ = 180°
32° + 27° + ∠YOZ = 180°
∠YOZ = 180° – 32° – 27° = 121°
Hence ∠YOZ = 121°, ∠OZY = 32°
- In Fig. 6.41, if AB || DE, ∠ BAC = 35° and ∠ CDE = 53°, find ∠ DCE.
Solution:
Let AB||DE , ∠ BAC = 35° and ∠ CDE = 53°
∠ BAC = ∠CED = 35° [alternate angles]
∠CED +∠CDE + ∠ECD = 180°
∠ECD = 180° – 53° – 35° = 92°
∠ECD = 92°
- In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠ TSQ = 75°, find ∠SQT
Solution:
Given PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠ TSQ = 75°,we have to find ∠SQT
∠ RPT + ∠ PRT + ∠ RTP = 180° [Sukm of interior angles is 180°]
∠ RTP = 180° – 95° – 40° = 45°
∠ STQ = ∠ RTP = 45° [vertically opposite angles]
In ΔSTQ , ∠ STQ + ∠ TSQ + ∠ SQT = 180°
75° + 45° + ∠ SQT = 180°
∠ SQT = 60°
- In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠ SQR = 28° and ∠ QRT = 65°, then find the values of x and y.
Solution:
PQ ⊥ PS, PQ || SR, ∠ SQR = 28° and ∠ QRT = 65°
∠PQS + ∠SQR = ∠QST
x + 28° = 65°
x = 65° – 28° = 37°
In ΔPQS, ∠QPS + ∠PSQ + ∠PQS = 180° [Sum of interior angles is 180°]
90° + 37° + ∠PSQ = 180°
∠PSQ = 53°
Therefore, x = 37° and y = 53°
6. In Fig. 6.44, the side QR of ∆ PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR.
Solution:
By exterior angle property, ∠PRS = ∠PQR + ∠QPR
Therefore, 1/2∠PRS = 1/2∠PQR + 1/2∠QPR
i.e., ∠TRS = ∠TQR + 1/2 ∠QPR……………..(1)
In ∆QTR, by exterior angle property
∠TRS = ∠TQR + ∠QTR ……………….(2)
From (1) and (2)
∠TQR + ∠QTR = ∠TQR + 1/2 ∠QPR
∠QTR =1/2 ∠QPR
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