In the figure, not drawn to scale. YZAB is a square. ABC is an equilateral triangle and YA is a straight line. Find
- One-third of ∠YDC
- Seven times of ∠CAD
(a)
∠BYA = 45° (Right angle)
∠DBA = 60° (Equilateral triangle)
∠YBD
= 90° - ∠DBA
= 90° - 60°
= 30°
∠YDC
= 30° + 45° (Exterior angle of a triangle)
= 75°
One-third of ∠YDC
=
13 x 75°
= 25°
(b)
∠CDA
= 180° - ∠YDC
= 180° - 75°
= 105° (Angles on a straight line)
∠CAD
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Seven times of ∠CAD
= 7 x 15°
= 105°
Answer(s): (a) 25°; (b) 105°
