(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|.
(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|.
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Correct Answer: Option n
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)
\(\frac{y}{6} = \frac{2}{5}\)
\(\therefore 5y = 2(6) = 12\)
\(\therefore y = \frac{12}{5} = 2.4 cm\)
\(\therefore XQ + QZ = XZ\)
\(QZ = XZ - XQ\)
= \(6 cm - 2.4 cm = 3.6 cm\)
\(\therefore QZ = 3.6 cm\).
(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)
\(\frac{y}{6} = \frac{2}{5}\)
\(\therefore 5y = 2(6) = 12\)
\(\therefore y = \frac{12}{5} = 2.4 cm\)
\(\therefore XQ + QZ = XZ\)
\(QZ = XZ - XQ\)
= \(6 cm - 2.4 cm = 3.6 cm\)
\(\therefore QZ = 3.6 cm\).