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 [tex]Inscribed angle=\frac{1}{2}intercepted arc[/tex], 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.
