In the figure given below, PQRS is a rhombus, SQ and PR are the diagonals of the rhombus
intersecting at O.
If angle OPQ= 35°, then find the value of angle ORS + angle OQP.
Answers
Answer:
The required sum = 90^o90
o
Step-by-step explanation:
We know that the diagonals of rhombus intersect at right angle.
So, \angle POQ = 90^o∠POQ=90
o
(refer the figure)
We can understand that;
\angle OPQ + \angle POQ + \angle OQP = 180^o∠OPQ+∠POQ+∠OQP=180
o
(refer the figure)
35^o + 90^o + \angle OQP = 180^o35
o
+90
o
+∠OQP=180
o
(refer the figure)
\angle OQP = 180^o - 125^o = 55^o∠OQP=180
o
−125
o
=55
o
(refer the figure)
Since, in the rhombus, the opposite sides are parallel.
Therefore, PQ is parallel to RS and PR is transversal line. (refer the figure)
So, \angle OPQ = \angle ORS∠OPQ=∠ORS {they are internal alternate angles.}
\angle OPQ = \angle ORS = 35^o∠OPQ=∠ORS=35
o
Hence, \angle ORS + \angle OQP = 35^o + 55^o = 90^o∠ORS+∠OQP=35
o
+55
o
=90
o
The required sum = 90^o90
o
Step-by-step explanation:
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Answer:
Value of angle ORS is 53°
because the sum of 2 angles of rohmbus is 90°