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Thanks Waterflow for your enlightenment!

0 Replies 0 Likes ✔Accepted Answer

Since ABHD is a trapezium, AH is parallel to BD.

If ∠DHG = u , ∠DHG = ∠HDB = u                 Alternate angles equal.

∠GDH = ∠DHG = u               triangle DGH is isosceles

∠DGF = ∠DHG + ∠GDH  = 2u                sum of interior angles equal exterior angle

∠FDG = 180º – 2* ∠DGF = 180º – 4u       triangle DFG is isosceles

 

Consider straight line EDC, all angles at D adds up to 180º

43º + 31º +∠HDB + ∠DHG + ∠FDG  = 180º

74º  + u + u + 180º – 4u = 180º

2u = 74º

u = 37º or ∠DHG = 37º

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Thanks Waterflow for your enlightenment!

0 Replies 0 Likes ✔Accepted Answer