In figure- ABparallel CDparallel EF- PQparallel RS- angle RQD-25-o and angle CQP-60-o- then angle QRS is equal to85-o135-o145-o110-o

Question

Toppr Admin · Accepted Answer

Given-xA0-RQD-25o-xA0-and-xA0-CQP-60oRQD-QRA-250-xA0-xA0-Alternate interior angles-CQP-BRS-600-Alternate Exterior angles-BRS-ARS-1800-xA0-Linear Pair-gt-ARS-1800-x2212-BRS-1800-x2212-600-1200-xA0-QRS-ARS-QRA-1200-250-1450-xA0-

In figure, if $A B ∥ C D ∥ E F, P Q ∥ R S, ∠ R Q D = 25^{o}$ and $∠ C Q P = 60^{o}$ , then $∠ Q R S$ is equal to

$85^{o}$
$135^{o}$
$145^{o}$
$110^{o}$

Given, RQD= $25^{o}$ and CQP= $60^{o}$
$R Q D = Q R A = 25^{0}$ (Alternate interior angles)
$C Q P = B R S = 60^{0}$ (Alternate Exterior angles)
$B R S + A R S = 180^{0}$ (Linear Pair)
$=> A R S = 180^{0} - B R S = 180^{0} - 60^{0} = 120^{0}$
$Q R S = A R S + Q R A = 120^{0} + 25^{0} = 145^{0}$

In figure, if AB∥CD∥EF,PQ∥RS,∠RQD=25o and ∠CQP=60o, then ∠QRS is equal to85o135o145o110o

Given, RQD=25o and CQP=60oRQD=QRA=250 (Alternate interior angles)CQP=BRS=600(Alternate Exterior angles)BRS+ARS=1800 (Linear Pair)=>ARS=1800−BRS=1800−600=1200 QRS=ARS+QRA=1200+250=1450

In figure, if $A B ∥ C D ∥ E F, P Q ∥ R S, ∠ R Q D = 25^{o}$ and $∠ C Q P = 60^{o}$ , then $∠ Q R S$ is equal to

$85^{o}$
$135^{o}$
$145^{o}$
$110^{o}$

Given, RQD= $25^{o}$ and CQP= $60^{o}$
$R Q D = Q R A = 25^{0}$ (Alternate interior angles)
$C Q P = B R S = 60^{0}$ (Alternate Exterior angles)
$B R S + A R S = 180^{0}$ (Linear Pair)
$=> A R S = 180^{0} - B R S = 180^{0} - 60^{0} = 120^{0}$
$Q R S = A R S + Q R A = 120^{0} + 25^{0} = 145^{0}$