Answer:
The angles are m∠EQF=14°, m∠EFQ=90° and m∠QEF=76°
Step-by-step explanation:
Given the figure in which m∠DFQ = 10°, m∠EOF = 28°
we have to find the angles of △EFQ
m∠EOF = 28°
By theorem angle subtended at the centre is twice the angle subtended at the circumference
⇒ m∠EQF=14°
As angle ∠EFQ is the angle in the semicircle
∴ ∠EFQ=90°
In ΔEFQ, by angle sum property of triangle
m∠EFQ+m∠FQE+m∠QEF=180°
90°+14°+m∠QEF=180°
m∠QEF=180°-90°-14°=76°
Hence, m∠EFQ=90° and m∠QEF=76°
