Answer:
m∠KEP=90°, m∠NPM=16°, m∠KPE=16° , m∠EKP=74°, m∠PMN=148°, m∠MNP=16°
Explanation:
Given In a rhombus MPKN with an obtuse angle K the diagonals intersect each other at point E.
The measure of one of the angles of a triangle PKE is equal 16°. we know diagonals of rhombus bisect at right angles i.e at 90°
Hence, m∠KEP=90°
As ∠K is given as obtuse angles so the only possible angle which measures 16° in triangle PKE is ∠KPE. Hence, m∠KPE=16°
In triangle KEP, by angle sum property
m∠EKP+m∠KEP+m∠KPE=180°
⇒ m∠EKP+90°+16°=180°
⇒ m∠EKP=74°
∵ The diagonals bisect the angles at vertices of rhombus.
⇒ ∠NPK=∠NPM=16°
and ∠PKN=2(∠PKE)=2(74°)=148°
∵ Opposite angles of rhombus are equal
⇒ ∠PKN=∠PMN=148°
In triangle PMN, By angle sum property
m∠NPM+m∠PMN+m∠MNP=180°
⇒ 16°+148°+m∠MNP=180°
⇒ m∠MNP=16°