(a) In the diagram, TU is tangent to the circle. < RVU = 100° and < URS = 36°. Calculate the value of angle STU.
(b) In triangle XYZ, |XY| = 5 cm, |YZ| = 8 cm and |XZ| = 6 cm. P is a point on the side XY such that |XP| = 2 cm and the line through P, parallel to YZ meets XZ at Q. Calculate |QZ|.
Explanation
(a) < UST = 100° (exterior angle of a cyclic quad)
< SUT = 36° (angle in alternate segment)
(therefore < UST + < SUT + < STU = 180°) (sum of angles in a triangle)
(therefore 100° + 36° + < STU = 180°)
(< STU = 180° – 136° = 44°)
(b)
(frac{y}{6} = frac{2}{5})
(therefore 5y = 2(6) = 12)
(therefore y = frac{12}{5} = 2.4 cm)
(therefore XQ + QZ = XZ)
(QZ = XZ – XQ)
= (6 cm – 2.4 cm = 3.6 cm)
(therefore QZ = 3.6 cm).