Advertisements
Advertisements
Question
In parallelogram LOST, SN ⊥ OL and SM ⊥ LT. Find ∠STM, ∠SON and ∠NSM.

Advertisements
Solution
Given, ∠MST = 40°
In ΔMST,
By the angle sum property of a triangle,
∠TMS + ∠MST + ∠STM = 180°
⇒ ∠STM = 180° – (90° + 40°) ...[∵ SM ⊥ LT, ∠TMS = 90°]
= 50°
∴ ∠SON = ∠STM = 50° ...[∵ Opposite angles of a parallelogram are equal]
Now, In the ΔONS,
∠ONS + ∠OSN + ∠SON = 180° ...[Angle sum property of triangle]
∠OSN = 180° – (90° + 50°)
= 180° – 140°
= 40°
Moreover, ∠SON + ∠TSO = 180° ...[∵ Adjacent angles of a parallelogram are supplementary]
⇒ ∠SON + ∠TSM + ∠NSM + ∠OSN = 180°
⇒ 50° + 40° + ∠NSM + 40° = 180°
⇒ 90° + 40° + ∠NSM = 180°
⇒ 130° + ∠NSM = 180°
⇒ ∠NSM = 180° – 130° = 50°
APPEARS IN
RELATED QUESTIONS
Can a quadrilateral ABCD be a parallelogram if ∠D + ∠B = 180°?
Can a quadrilateral ABCD be a parallelogram if ∠A = 70° and ∠C = 65°?
In the given figure, `square`PQRS and `square`ABCR are two parallelograms. If ∠P = 110° then find the measures of all angles of `square`ABCR.

ABCD is a parallelogram. What kind of quadrilateral is it if: AC = BD but AC is not perpendicular to BD?
Prove that the diagonals of a parallelogram bisect each other.
In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that:
- AE = AD,
- DE bisects and ∠ADC and
- Angle DEC is a right angle.
All rectangles are parallelograms.
In parallelogram ABCD, the angle bisector of ∠A bisects BC. Will angle bisector of B also bisect AD? Give reason.
In the following figure, FD || BC || AE and AC || ED. Find the value of x.

Construct a parallelogram ABCD in which AB = 4 cm, BC = 5 cm and ∠B = 60°.
