In the figure, ∠BGL is a right-angled isosceles triangle. BL // DK , ∠JFE = 50°, ∠GJH = 42° and ∠EHF = 57°. Find
- ∠BLG
- ∠FDK
- ∠HKJ
(a)
∠BLG
= (180° - 90°) ÷ 2
= 90° ÷ 2
= 45° (Isosceles triangle)
(b)
∠GJC = 45° (Corresponding angles, LB//JC)
∠FJD
= ∠GJH + ∠GJC
= 42° + 45°
= 87°
∠FDK
= 180° - ∠JFE - ∠FJD
= 180° - 50° - 87°
= 43° (Angles sum of triangle)
(c)
∠HJK
= 180° - ∠FJD
= 180° - 87°
= 93°(Angles in a straight line)
∠KHJ = ∠FHG = 57° (Vertically opposite angles)
∠HKJ
= 180° - ∠KHJ - ∠HJK
= 180° - 57° - 93°
= 30° (Angles sum of triangle)
Answer(s): (a) 45°; (b) 43°; (c) 30°
