Answer:
x°=70° and y°=40°
Step-by-step explanation:
We are given that secants PS and PT and tangent QR intersecting at point P. Triangle STP is an isosceles triangle with legs PS and PT.
Now, we know that \(Inscribed angle=\frac{1}{2}intercepted arc\), thus
⇒2m∠PST=m(TP)
⇒2m∠PST=140°
⇒m∠PST=70°
Since, the ΔPST is an isosceles triangle with SP=TP, therefore
∠PST=∠PTS=70°
Now,in ΔPST, applying the angle sum property, we have
∠PST+∠PTS+∠SPT=180°
⇒70°+70°+∠SPT=180°
⇒140°+∠SPT=180°
⇒∠SPT=180°-140°
⇒∠SPT=40°
therefore, y°=40°
Now, using the theorem that angle made with the tangent is equal to the interior angle of the inscibed triangle, thus
x°=∠PTS
⇒x°=70°
Thus, the value of x and y is: 70 and 40 respectively.
