In the figure, not drawn to scale. NPQR is a square. QRS is an equilateral triangle and NQ is a straight line. Find
- Three-fifths of ∠NTS
- Six times of ∠SQT
(a)
∠RNQ = 45° (Right angle)
∠TRQ = 60° (Equilateral triangle)
∠NRT
= 90° - ∠TRQ
= 90° - 60°
= 30°
∠NTS
= 30° + 45° (Exterior angle of a triangle)
= 75°
Three-fifths of ∠NTS
=
35 x 75°
= 45°
(b)
∠STQ
= 180° - ∠NTS
= 180° - 75°
= 105° (Angles on a straight line)
∠SQT
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Six times of ∠SQT
= 6 x 15°
= 90°
Answer(s): (a) 45°; (b) 90°