angle QRS = 51°
<h2>Step-by-step explanation:</h2>
The diagram for the question has been attached to this response.
From the diagram, some circle theorems can be applied.
<em><u>Circle theorem:</u></em>
The angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any other point on the remaining part of the circle.
From the diagram, QR is the arc and therefore:
∠QSR = 2 x ∠ QTR
<em>Since ∠QTR = 39°</em>
=> ∠QSR = 2 x 39°
=> ∠QSR = 78°
<u><em>Other theorem:</em></u>
(i)The base angles of an isosceles triangle are equal.
Triangle QSR is an isosceles triangle, therefore, angles SQR and QRS are equal. i.e
∠SQR = ∠QRS
(ii) The sum of angles of a triangle is 180°. i.e
∠SQR + ∠QRS + ∠QSR = 180°
<em>Since ∠SQR = ∠QRS and ∠QSR = 78°</em>
=> ∠QRS + ∠QRS + 78° = 180°
=> 2(∠QRS) = 180° - 78°
=> 2(∠QRS) = 102°
<em>Divide both sides by 2</em>
=> ∠QRS = 51°
Therefore, angle QRS = 51°
