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Question
primary 6 | Maths
| Geometry
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Date Posted: 5 years ago
Views: 284
(a)
∡HJK + ∡JKB + ∡AKB = 180° &
∡JKB = ∡AKB
so ∡JKB = (180-48)/2 = 66°
&
∡KBJ = 90-66 = 24°
∡KBJ = ∡ABK = 24°
∡JBC + ∡KBJ + ∡ABK = 180°
thus,
∡JBC = 180-24-24 = 132°
(b)
∡ECF = ∡EFC = 45°
∡DCE = ∡DFE = 20°
&
∡DCE + ∡DFE + ∡EFC + ∡ECF + ∡CDF = 180°
thus,
∡CDF = 180-45-45-20-20 = 50°
∡HJK + ∡JKB + ∡AKB = 180° &
∡JKB = ∡AKB
so ∡JKB = (180-48)/2 = 66°
&
∡KBJ = 90-66 = 24°
∡KBJ = ∡ABK = 24°
∡JBC + ∡KBJ + ∡ABK = 180°
thus,
∡JBC = 180-24-24 = 132°
(b)
∡ECF = ∡EFC = 45°
∡DCE = ∡DFE = 20°
&
∡DCE + ∡DFE + ∡EFC + ∡ECF + ∡CDF = 180°
thus,
∡CDF = 180-45-45-20-20 = 50°
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Date Posted:
5 years ago
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