In the figure, not drawn to scale. HJKL is a square. KLM is an equilateral triangle and HK is a straight line. Find
- One-third of ∠HNM
- Seven times of ∠MKN
(a)
∠LHK = 45° (Right angle)
∠NLK = 60° (Equilateral triangle)
∠HLN
= 90° - ∠NLK
= 90° - 60°
= 30°
∠HNM
= 30° + 45° (Exterior angle of a triangle)
= 75°
One-third of ∠HNM
=
13 x 75°
= 25°
(b)
∠MNK
= 180° - ∠HNM
= 180° - 75°
= 105° (Angles on a straight line)
∠MKN
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Seven times of ∠MKN
= 7 x 15°
= 105°
Answer(s): (a) 25°; (b) 105°
