Question

in rectangle PQRS diagonals intersect at o . if angle RPQ=30degree angleRQO=x and angleROS=y. find the value of x+y

Answer 1

angle RPQ = 30°
angle QPS= 90°

angle RPS =QPS - RPQ
= 90 - 30
= 60°
angle PQS = angle RPQ = 30°
angle x = angle RPS = 60°
angle QSR = angle QPR = 30°
angle QSR = angle PRS = 30°

Sum of angles in Triangle = 180°
In ∆OSR
30° + 30°+ y = 180°
y = 180 -60 = 120°


Therefore x+y = 60° + 120° = 180°

Answer 2

Given:

Angle RPQ=30°

Angle RQO=x

Angle ROS=y

To find:

The value of x+y

Solution:

The value of x+y is 180°.

We can find the value by following the given steps-

We know that PQRS is a rectangle and its diagonals bisect at O.

Since its diagonals are also equal, we get PO=OQ.

Now, in ΔPOQ, PO=OQ.

So, angle OPQ=angle OQP=30° (angles opposite to equal sides in a triangle)

Also, angle POQ+angle OPQ+ angle OQP=180°

Using the values,

Angle POQ+30°+30°=180°

Angle POQ+60°=180°

Angle POQ=180°-60°

Angle POQ=120°

The lines PQ and RS are parallel to each other and PR and SQ intersect at O.

So, angle POQ=angle ROS as they are vertically opposite.

Angle ROS=y=120°

We know that the angle Q is 90°.

Angle Q=angle RQO+angle OQP

On putting the values,

90°=angle RQO+30°

Angle RQO=90°-30°

Angle RQO=x=60°

Now, we will add the angles RQO and ROS to obtain x+y.

x+y=angle RQO+angle ROS

x+y=60°+120°

x=y=180°

Therefore, the value of x+y is 180°.