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Draw BE⊥PQ and CF⊥RS. ∠1 = ∠2 ...(i)[∵ Angle of incident = Angle of reflection] Similarly, ∠3 = ∠4 ...(ii) and, ∠2 = ∠3 ...(iii)[∵ Alternate Angles] ⇒ ∠1 = ∠4 [From the equations (i), (ii) and (iii)] ⇒ 2∠1 = 2∠4 ⇒ ∠1 + ∠1 = ∠4 + ∠4 ⇒ ∠1 + ∠2 = ∠3 + ∠4 [From the equation (i) and (ii)] ⇒ ∠BCD = ∠ABC SRead more
Draw BE⊥PQ and CF⊥RS. ∠1 = ∠2 …(i)[∵ Angle of incident = Angle of reflection] Similarly, ∠3 = ∠4 …(ii) and, ∠2 = ∠3 …(iii)[∵ Alternate Angles] ⇒ ∠1 = ∠4 [From the equations (i), (ii) and (iii)] ⇒ 2∠1 = 2∠4 ⇒ ∠1 + ∠1 = ∠4 + ∠4 ⇒ ∠1 + ∠2 = ∠3 + ∠4 [From the equation (i) and (ii)] ⇒ ∠BCD = ∠ABC Since, the alternate angles are equal. Hence, AB||CD.
Given that: ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75° In △PTR, ∠P + ∠R + ∠PTR = 180° ⇒ 95° + 40° + ∠PTR = 180° ⇒ 135° + ∠PTR = 180° ⇒ ∠PTR = 180° - 135° = 45° ∠STQ = ∠PTR [∵ Vertically Opposite Angles] ⇒ ∠STQ = 45° In △SQT, ∠STQ + ∠S + ∠SQT = 180° ⇒ 45° + 75° + ∠SQT = 180° ⇒ 120° + ∠SQT = 180° ⇒ ∠SQT =Read more
Given that: ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75° In △PTR, ∠P + ∠R + ∠PTR = 180° ⇒ 95° + 40° + ∠PTR = 180° ⇒ 135° + ∠PTR = 180° ⇒ ∠PTR = 180° – 135° = 45° ∠STQ = ∠PTR [∵ Vertically Opposite Angles] ⇒ ∠STQ = 45° In △SQT, ∠STQ + ∠S + ∠SQT = 180° ⇒ 45° + 75° + ∠SQT = 180° ⇒ 120° + ∠SQT = 180° ⇒ ∠SQT = 180° – 120° = 60°
∠PRS is the exterior angle of △PQR. Therefore, ∠PRS = ∠QPR + ∠PQR ⇒ 1/2 ∠PRS = 1/2 ∠QPR + 1/2 ∠PQR ⇒ ∠TRS = 1/2 ∠QPR + ∠TQR ...(1)[∵ ∠TRS = 1/2 ∠PRS and ∠TQR = 1/2 ∠PQR] ∠TRS is exterior angle of △TQR Therefore, ∠TRS = ∠QTR + ∠TQR ...(2) From, the equations (1) and (2), we have ∠QTR + ∠TQR = 1/2 ∠QPRead more
∠PRS is the exterior angle of △PQR. Therefore, ∠PRS = ∠QPR + ∠PQR ⇒ 1/2 ∠PRS = 1/2 ∠QPR + 1/2 ∠PQR ⇒ ∠TRS = 1/2 ∠QPR + ∠TQR …(1)[∵ ∠TRS = 1/2 ∠PRS and ∠TQR = 1/2 ∠PQR] ∠TRS is exterior angle of △TQR Therefore, ∠TRS = ∠QTR + ∠TQR …(2) From, the equations (1) and (2), we have ∠QTR + ∠TQR = 1/2 ∠QPR + ∠TQR ⇒ ∠QTR = 1/2 ∠QPR
Given that: PQ⊥PS||SR, ∠SQR = 28° and ∠QRT = 65° ∠PQR = ∠QRT [∵ Alternate Angles] ⇒ ∠RQS + ∠PQS = 65° ⇒ 28° + x = 65° ⇒ x = 65° - 28° = 37° In △PQS, ∠P + ∠PQS + ∠PSQ = 180° ⇒ 90° + 37° + y = 180° ⇒ 127° + y = 180° ⇒ y = 180° - 127° = 53°
Given that: PQ⊥PS||SR, ∠SQR = 28° and ∠QRT = 65° ∠PQR = ∠QRT [∵ Alternate Angles] ⇒ ∠RQS + ∠PQS = 65° ⇒ 28° + x = 65° ⇒ x = 65° – 28° = 37° In △PQS, ∠P + ∠PQS + ∠PSQ = 180° ⇒ 90° + 37° + y = 180° ⇒ 127° + y = 180° ⇒ y = 180° – 127° = 53°
Given that: PQ||ST, Therefore, ∠PQR = ∠APQ [∵ Alternate Angles] ⇒ x = 50° ∠APR = ∠PRD [∵ Alternate Angles] ⇒ ∠APQ + ∠QPR = 127° ⇒ 50° + y = 127° ⇒ y = 127° - 50° = 77°
Given that: PQ||ST, Therefore, ∠PQR = ∠APQ [∵ Alternate Angles] ⇒ x = 50° ∠APR = ∠PRD [∵ Alternate Angles] ⇒ ∠APQ + ∠QPR = 127° ⇒ 50° + y = 127° ⇒ y = 127° – 50° = 77°
In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other.
Draw BE⊥PQ and CF⊥RS. ∠1 = ∠2 ...(i)[∵ Angle of incident = Angle of reflection] Similarly, ∠3 = ∠4 ...(ii) and, ∠2 = ∠3 ...(iii)[∵ Alternate Angles] ⇒ ∠1 = ∠4 [From the equations (i), (ii) and (iii)] ⇒ 2∠1 = 2∠4 ⇒ ∠1 + ∠1 = ∠4 + ∠4 ⇒ ∠1 + ∠2 = ∠3 + ∠4 [From the equation (i) and (ii)] ⇒ ∠BCD = ∠ABC SRead more
Draw BE⊥PQ and CF⊥RS.
See less∠1 = ∠2 …(i)[∵ Angle of incident = Angle of reflection]
Similarly,
∠3 = ∠4 …(ii)
and,
∠2 = ∠3 …(iii)[∵ Alternate Angles]
⇒ ∠1 = ∠4 [From the equations (i), (ii) and (iii)]
⇒ 2∠1 = 2∠4
⇒ ∠1 + ∠1 = ∠4 + ∠4
⇒ ∠1 + ∠2 = ∠3 + ∠4 [From the equation (i) and (ii)]
⇒ ∠BCD = ∠ABC
Since, the alternate angles are equal. Hence, AB||CD.
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 that: ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75° In △PTR, ∠P + ∠R + ∠PTR = 180° ⇒ 95° + 40° + ∠PTR = 180° ⇒ 135° + ∠PTR = 180° ⇒ ∠PTR = 180° - 135° = 45° ∠STQ = ∠PTR [∵ Vertically Opposite Angles] ⇒ ∠STQ = 45° In △SQT, ∠STQ + ∠S + ∠SQT = 180° ⇒ 45° + 75° + ∠SQT = 180° ⇒ 120° + ∠SQT = 180° ⇒ ∠SQT =Read more
Given that: ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°
See lessIn △PTR, ∠P + ∠R + ∠PTR = 180°
⇒ 95° + 40° + ∠PTR = 180°
⇒ 135° + ∠PTR = 180°
⇒ ∠PTR = 180° – 135° = 45°
∠STQ = ∠PTR [∵ Vertically Opposite Angles]
⇒ ∠STQ = 45°
In △SQT, ∠STQ + ∠S + ∠SQT = 180°
⇒ 45° + 75° + ∠SQT = 180°
⇒ 120° + ∠SQT = 180°
⇒ ∠SQT = 180° – 120° = 60°
In Fig. 6.44, the side QR of ∆ PQR is produced to a point S.
∠PRS is the exterior angle of △PQR. Therefore, ∠PRS = ∠QPR + ∠PQR ⇒ 1/2 ∠PRS = 1/2 ∠QPR + 1/2 ∠PQR ⇒ ∠TRS = 1/2 ∠QPR + ∠TQR ...(1)[∵ ∠TRS = 1/2 ∠PRS and ∠TQR = 1/2 ∠PQR] ∠TRS is exterior angle of △TQR Therefore, ∠TRS = ∠QTR + ∠TQR ...(2) From, the equations (1) and (2), we have ∠QTR + ∠TQR = 1/2 ∠QPRead more
∠PRS is the exterior angle of △PQR.
See lessTherefore,
∠PRS = ∠QPR + ∠PQR
⇒ 1/2 ∠PRS = 1/2 ∠QPR + 1/2 ∠PQR
⇒ ∠TRS = 1/2 ∠QPR + ∠TQR …(1)[∵ ∠TRS = 1/2 ∠PRS and ∠TQR = 1/2 ∠PQR]
∠TRS is exterior angle of △TQR
Therefore,
∠TRS = ∠QTR + ∠TQR …(2)
From, the equations (1) and (2), we have
∠QTR + ∠TQR = 1/2 ∠QPR + ∠TQR
⇒ ∠QTR = 1/2 ∠QPR
In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠ SQR = 28° and ∠ QRT = 65°, then find the values of x and y.
Given that: PQ⊥PS||SR, ∠SQR = 28° and ∠QRT = 65° ∠PQR = ∠QRT [∵ Alternate Angles] ⇒ ∠RQS + ∠PQS = 65° ⇒ 28° + x = 65° ⇒ x = 65° - 28° = 37° In △PQS, ∠P + ∠PQS + ∠PSQ = 180° ⇒ 90° + 37° + y = 180° ⇒ 127° + y = 180° ⇒ y = 180° - 127° = 53°
Given that: PQ⊥PS||SR, ∠SQR = 28° and ∠QRT = 65°
See less∠PQR = ∠QRT [∵ Alternate Angles]
⇒ ∠RQS + ∠PQS = 65°
⇒ 28° + x = 65°
⇒ x = 65° – 28° = 37°
In △PQS, ∠P + ∠PQS + ∠PSQ = 180°
⇒ 90° + 37° + y = 180°
⇒ 127° + y = 180°
⇒ y = 180° – 127° = 53°
In Fig. 6.32, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.
Given that: PQ||ST, Therefore, ∠PQR = ∠APQ [∵ Alternate Angles] ⇒ x = 50° ∠APR = ∠PRD [∵ Alternate Angles] ⇒ ∠APQ + ∠QPR = 127° ⇒ 50° + y = 127° ⇒ y = 127° - 50° = 77°
Given that: PQ||ST,
See lessTherefore,
∠PQR = ∠APQ [∵ Alternate Angles]
⇒ x = 50°
∠APR = ∠PRD [∵ Alternate Angles]
⇒ ∠APQ + ∠QPR = 127°
⇒ 50° + y = 127°
⇒ y = 127° – 50° = 77°