Question

PQRS is a rectangle. The diagonals intersect at O. Diagonal RP is produced to T. ∠QPT = 142°. Find the angles of ΔSOR.

Answer 1

Given:

• PQRS is a rectangle.
• Diagonals intersect at O (they bisect each other).
• Diagonal RP is extended to T.
• ∠QPT = 142°

Determine ∠QPR:

Since ∠QPT = 142° and ∠QPR and ∠QPT form a linear pair on the line RT, their sum is 180°.

Therefore, ∠QPR = 180° – 142° = 38°.

Determine ∠PRS:

In a rectangle, opposite sides are parallel, so PQ || SR.

Since PR is a transversal, ∠QPR and ∠PRS are alternate interior angles.

Therefore, ∠PRS = ∠QPR = 38°.

Determine ∠PSR:

In a rectangle, all interior angles are right angles.

Therefore, ∠PSR = 90°.

Determine ∠SPR:

In right-angled triangle PSR, the sum of angles is 180°.

Therefore, ∠SPR = 180° – ∠PSR – ∠PRS

= 180° – 90° – 38°

= 52°

Determine ∠OSR: 

Since the diagonals of a rectangle bisect each other and are equal in length, OS = OR.

Therefore, ΔSOR is an isosceles triangle and the base angles opposite to the equal sides are equal.

Thus, ∠OSR = ∠ORS = ∠PRS = 38°.

Determine ∠SOR: 

In ΔSOR, the sum of angles is 180°.

Therefore, ∠SOR = 180° – ∠OSR – ∠ORS

= 180° – 38° – 38°

= 180° – 76°

= 104°